{
  "index": "2007-A-1",
  "type": "ALG",
  "tag": [
    "ALG",
    "GEO",
    "NT"
  ],
  "difficulty": "",
  "question": "Find all values of $\\alpha$ for which the curves $y = \\alpha x^2 +\n\\alpha x + \\frac{1}{24}$ and $x = \\alpha y^2 + \\alpha y + \\frac{1}{24}$\nare tangent to each other.",
  "solution": "The only such $\\alpha$ are $2/3, 3/2, (13 \\pm \\sqrt{601})/12$.\n\n\\textbf{First solution:}\nLet $C_1$ and $C_2$ be the curves\n$y=\\alpha x^2 + \\alpha x + \\frac{1}{24}$\nand $x=\\alpha y^2 + \\alpha y + \\frac{1}{24}$, respectively,\nand let $L$ be the line $y=x$.\nWe consider three cases.\n\nIf $C_1$ is tangent to $L$, then the point of tangency $(x,x)$ satisfies\n\\[\n2\\alpha x + \\alpha = 1, \\qquad x = \\alpha x^2 + \\alpha x + \\frac{1}{24};\n\\]\nby symmetry, $C_2$ is tangent to $L$ there, so $C_1$ and $C_2$ are tangent.\nWriting $\\alpha = 1/(2x+1)$ in the first equation and substituting into\nthe second, we must have\n\\[\nx = \\frac{x^2+x}{2x+1} + \\frac{1}{24},\n\\]\nwhich simplifies to $0 = 24x^2 - 2x - 1\n= (6x+1)(4x-1)$, or $x \\in \\{1/4, -1/6\\}$. This yields\n$\\alpha = 1/(2x+1) \\in \\{2/3, 3/2\\}$.\n\nIf $C_1$ does not intersect $L$, then $C_1$ and $C_2$ are separated by $L$\nand so cannot be tangent.\n\nIf $C_1$ intersects $L$ in two distinct points $P_1, P_2$, then it is not\ntangent to $L$ at either point. Suppose at one of these points, say $P_1$,\nthe tangent to $C_1$ is perpendicular to $L$; then by symmetry, the same\nwill be true of $C_2$, so $C_1$ and $C_2$ will be tangent at $P_1$. In this\ncase, the point $P_1 = (x,x)$ satisfies\n\\[\n2 \\alpha x + \\alpha = -1, \\qquad x = \\alpha x^2 + \\alpha x + \\frac{1}{24};\n\\]\nwriting $\\alpha = -1/(2x+1)$ in the first equation and substituting into\nthe second, we have\n\\[\nx = -\\frac{x^2+x}{2x+1} + \\frac{1}{24},\n\\]\nor\n$x = (-23 \\pm \\sqrt{601})/72$.\nThis yields\n$\\alpha = -1/(2x+1) = (13 \\pm \\sqrt{601})/12$.\n\nIf instead the tangents to $C_1$ at $P_1, P_2$ are not perpendicular to\n$L$, then we claim there cannot be any point where $C_1$ and $C_2$ are tangent.\nIndeed, if we count intersections of $C_1$ and $C_2$ (by using $C_1$ to\nsubstitute for $y$ in $C_2$, then solving for $y$), we get at most four\nsolutions counting multiplicity. Two of these are $P_1$ and $P_2$,\nand any point of tangency counts for two more. However, off of $L$,\nany point of tangency would have a mirror image which is also a point of\ntangency, and there cannot be six solutions. Hence we have now found all\npossible $\\alpha$.\n\n\\textbf{Second solution:}\nFor any nonzero value of $\\alpha$, the two\nconics will intersect in four points in the complex projective plane\n$\\mathbb{P}^2(\\mathbb{C})$. To determine the\n$y$-coordinates of these intersection points, subtract the two equations\nto obtain\n\\[\n(y-x) = \\alpha(x-y)(x+y) + \\alpha(x-y).\n\\]\nTherefore, at\na point of intersection we have either $x=y$, or $x = -1/\\alpha - (y+1)$.\nSubstituting these two possible linear conditions into the second\nequation shows that the $y$-coordinate of a point of intersection is a\nroot of either\n$Q_1(y) = \\alpha y^2+(\\alpha-1)y + 1/24$ or\n$Q_2(y) = \\alpha y^2 + (\\alpha+1) y + 25/24 +1/\\alpha$.\n\nIf two curves\nare tangent, then the $y$-coordinates of at least two of the\nintersection points will coincide; the converse is also true because one of the\ncurves is the graph of a function in $x$. The coincidence\noccurs precisely when either\nthe discriminant of at least one of $Q_1$ or $Q_2$ is zero, or\nthere is a common root of $Q_1$ and $Q_2$. Computing the discriminants of $Q_1$\nand $Q_2$ yields (up to constant factors) $f_1(\\alpha)=6\\alpha^2 -\n13\\alpha + 6$ and $f_2(\\alpha)=6\\alpha^2 - 13\\alpha - 18$, respectively.\nIf on the other hand $Q_1$ and $Q_2$ have a common root, it must\nbe also a root of $Q_2(y) - Q_1(y) = 2y +1 + 1/\\alpha$,\nyielding $y = -(1+\\alpha)/(2\\alpha)$ and\n$0 = Q_1(y) = -f_2(\\alpha)/(24 \\alpha)$.\n\nThus the values of\n$\\alpha$ for which the two curves are tangent must be contained in the\nset of zeros of $f_1$ and $f_2$, namely $2/3$, $3/2$, and\n$(13\\pm\\sqrt{601})/12$.\n\n\\textbf{Remark:}\nThe fact that the two conics in $\\mathbb{P}^2(\\CC)$ meet in four points, counted\nwith multiplicities, is a special case of \\emph{B\\'ezout's theorem}: two\ncurves in $\\mathbb{P}^2(\\CC)$ of degrees $m, n$ and not sharing any common component\nmeet in exactly $mn$ points when counted with multiplicity.\n\nMany solvers were surprised that the proposers chose the parameter $1/24$\nto give two rational roots and two nonrational roots. In fact, they had\nno choice in the matter: attempting to make all four roots rational\nby replacing $1/24$ by $\\beta$ amounts\nto asking for $\\beta^2 + \\beta$ and $\\beta^2 + \\beta + 1$ to be perfect\nsquares. This cannot happen outside of trivial cases ($\\beta = 0, -1$)\nultimately because the elliptic curve 24A1 (in Cremona's notation)\nover $\\mathbb{Q}$ has rank $0$. (Thanks to Noam Elkies for providing this\ncomputation.)\n\nHowever, there are choices that make the radical milder,\ne.g., $\\beta = 1/3$  gives\n$\\beta^2 + \\beta = 4/9$ and $\\beta^2 + \\beta + 1 = 13/9$,\nwhile $\\beta = 3/5$ gives $\\beta^2 + \\beta = 24/25$\nand $\\beta^2 + \\beta + 1 = 49/25$.",
  "vars": [
    "x",
    "y",
    "L",
    "C_1",
    "C_2",
    "P_1",
    "P_2",
    "Q_1",
    "Q_2",
    "f_1",
    "f_2",
    "m",
    "n"
  ],
  "params": [
    "\\\\alpha",
    "\\\\beta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "L": "diagonal",
        "C_1": "firstcurve",
        "C_2": "secondcurve",
        "P_1": "firstpoint",
        "P_2": "secondpoint",
        "Q_1": "firstquad",
        "Q_2": "secondquad",
        "f_1": "firstpoly",
        "f_2": "secondpoly",
        "m": "degreeone",
        "n": "degreetwo",
        "\\alpha": "coefalpha",
        "\\beta": "coefbeta"
      },
      "question": "Find all values of $coefalpha$ for which the curves $ordinate = coefalpha abscissa^2 +\ncoefalpha abscissa + \\frac{1}{24}$ and $abscissa = coefalpha ordinate^2 + coefalpha ordinate + \\frac{1}{24}$\nare tangent to each other.",
      "solution": "The only such $coefalpha$ are $2/3, 3/2, (13 \\pm \\sqrt{601})/12$.\n\n\\textbf{First solution:}\nLet $firstcurve$ and $secondcurve$ be the curves\n$ordinate=coefalpha abscissa^2 + coefalpha abscissa + \\frac{1}{24}$\nand $abscissa=coefalpha ordinate^2 + coefalpha ordinate + \\frac{1}{24}$, respectively,\nand let $diagonal$ be the line $ordinate=abscissa$.\nWe consider three cases.\n\nIf $firstcurve$ is tangent to $diagonal$, then the point of tangency $(abscissa,abscissa)$ satisfies\n\\[\n2coefalpha abscissa + coefalpha = 1, \\qquad abscissa = coefalpha abscissa^2 + coefalpha abscissa + \\frac{1}{24};\n\\]\nby symmetry, $secondcurve$ is tangent to $diagonal$ there, so $firstcurve$ and $secondcurve$ are tangent.\nWriting $coefalpha = 1/(2abscissa+1)$ in the first equation and substituting into\nthe second, we must have\n\\[\nabscissa = \\frac{abscissa^2+abscissa}{2abscissa+1} + \\frac{1}{24},\n\\]\nwhich simplifies to $0 = 24abscissa^2 - 2abscissa - 1\n= (6abscissa+1)(4abscissa-1)$, or $abscissa \\in \\{1/4, -1/6\\}$. This yields\n$coefalpha = 1/(2abscissa+1) \\in \\{2/3, 3/2\\}$.\n\nIf $firstcurve$ does not intersect $diagonal$, then $firstcurve$ and $secondcurve$ are separated by $diagonal$\nand so cannot be tangent.\n\nIf $firstcurve$ intersects $diagonal$ in two distinct points $firstpoint, secondpoint$, then it is not\ntangent to $diagonal$ at either point. Suppose at one of these points, say $firstpoint$,\nthe tangent to $firstcurve$ is perpendicular to $diagonal$; then by symmetry, the same\nwill be true of $secondcurve$, so $firstcurve$ and $secondcurve$ will be tangent at $firstpoint$. In this\ncase, the point $firstpoint = (abscissa,abscissa)$ satisfies\n\\[\n2 \\, coefalpha \\, abscissa + coefalpha = -1, \\qquad abscissa = coefalpha abscissa^2 + coefalpha abscissa + \\frac{1}{24};\n\\]\nwriting $coefalpha = -1/(2abscissa+1)$ in the first equation and substituting into\nthe second, we have\n\\[\nabscissa = -\\frac{abscissa^2+abscissa}{2abscissa+1} + \\frac{1}{24},\n\\]\nor\n$abscissa = (-23 \\pm \\sqrt{601})/72$.\nThis yields\n$coefalpha = -1/(2abscissa+1) = (13 \\pm \\sqrt{601})/12$.\n\nIf instead the tangents to $firstcurve$ at $firstpoint, secondpoint$ are not perpendicular to\n$diagonal$, then we claim there cannot be any point where $firstcurve$ and $secondcurve$ are tangent.\nIndeed, if we count intersections of $firstcurve$ and $secondcurve$ (by using $firstcurve$ to\nsubstitute for $ordinate$ in $secondcurve$, then solving for $ordinate$), we get at most four\nsolutions counting multiplicity. Two of these are $firstpoint$ and $secondpoint$,\nand any point of tangency counts for two more. However, off of $diagonal$,\nany point of tangency would have a mirror image which is also a point of\ntangency, and there cannot be six solutions. Hence we have now found all\npossible $coefalpha$.\n\n\\textbf{Second solution:}\nFor any nonzero value of $coefalpha$, the two\nconics will intersect in four points in the complex projective plane\n$\\mathbb{P}^2(\\mathbb{C})$. To determine the\n$ordinate$-coordinates of these intersection points, subtract the two equations\nto obtain\n\\[\n(ordinate-abscissa) = coefalpha(abscissa-ordinate)(abscissa+ordinate) + coefalpha(abscissa-ordinate).\n\\]\nTherefore, at\na point of intersection we have either $abscissa=ordinate$, or $abscissa = -1/coefalpha - (ordinate+1)$.\nSubstituting these two possible linear conditions into the second\nequation shows that the $ordinate$-coordinate of a point of intersection is a\nroot of either\n$firstquad(ordinate) = coefalpha ordinate^2+(coefalpha-1)ordinate + 1/24$ or\n$secondquad(ordinate) = coefalpha ordinate^2 + (coefalpha+1) ordinate + 25/24 +1/coefalpha$.\n\nIf two curves\nare tangent, then the $ordinate$-coordinates of at least two of the\nintersection points will coincide; the converse is also true because one of the\ncurves is the graph of a function in $abscissa$. The coincidence\noccurs precisely when either\nthe discriminant of at least one of $firstquad$ or $secondquad$ is zero, or\nthere is a common root of $firstquad$ and $secondquad$. Computing the discriminants of $firstquad$\nand $secondquad$ yields (up to constant factors) $firstpoly(coefalpha)=6\\,coefalpha^2 -\n13\\,coefalpha + 6$ and $secondpoly(coefalpha)=6\\,coefalpha^2 - 13\\,coefalpha - 18$, respectively.\nIf on the other hand $firstquad$ and $secondquad$ have a common root, it must\nbe also a root of $secondquad(ordinate) - firstquad(ordinate) = 2\\,ordinate +1 + 1/coefalpha$,\nyielding $ordinate = -(1+coefalpha)/(2\\,coefalpha)$ and\n$0 = firstquad(ordinate) = -secondpoly(coefalpha)/(24 \\, coefalpha)$.\n\nThus the values of\n$coefalpha$ for which the two curves are tangent must be contained in the\nset of zeros of $firstpoly$ and $secondpoly$, namely $2/3$, $3/2$, and\n$(13\\pm\\sqrt{601})/12$.\n\n\\textbf{Remark:}\nThe fact that the two conics in $\\mathbb{P}^2(\\CC)$ meet in four points, counted\nwith multiplicities, is a special case of \\emph{B\\'ezout's theorem}: two\ncurves in $\\mathbb{P}^2(\\CC)$ of degrees $degreeone, degreetwo$ and not sharing any common component\nmeet in exactly $degreeone degreetwo$ points when counted with multiplicity.\n\nMany solvers were surprised that the proposers chose the parameter $1/24$\nto give two rational roots and two nonrational roots. In fact, they had\nno choice in the matter: attempting to make all four roots rational\nby replacing $1/24$ by $coefbeta$ amounts\nto asking for $coefbeta^2 + coefbeta$ and $coefbeta^2 + coefbeta + 1$ to be perfect\nsquares. This cannot happen outside of trivial cases ($coefbeta = 0, -1$)\nultimately because the elliptic curve 24A1 (in Cremona's notation)\nover $\\mathbb{Q}$ has rank $0$. (Thanks to Noam Elkies for providing this\ncomputation.)\n\nHowever, there are choices that make the radical milder,\ne.g., $coefbeta = 1/3$  gives\n$coefbeta^2 + coefbeta = 4/9$ and $coefbeta^2 + coefbeta + 1 = 13/9$,\nwhile $coefbeta = 3/5$ gives $coefbeta^2 + coefbeta = 24/25$\nand $coefbeta^2 + coefbeta + 1 = 49/25$."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "giraffewing",
        "y": "pumpkinlace",
        "L": "harvestmoon",
        "C_1": "butterflynet",
        "C_2": "lanternlight",
        "P_1": "whistlegrass",
        "P_2": "thimbleberry",
        "Q_1": "cobblestone",
        "Q_2": "dandelionseed",
        "f_1": "caterpillar",
        "f_2": "dragonfly",
        "m": "storyteller",
        "n": "marshmallow",
        "\\\\alpha": "porcelain",
        "\\\\beta": "blackpepper"
      },
      "question": "Find all values of $porcelain$ for which the curves $pumpkinlace = porcelain giraffewing^2 +\nporcelain giraffewing + \\frac{1}{24}$ and $giraffewing = porcelain pumpkinlace^2 + porcelain pumpkinlace + \\frac{1}{24}$\nare tangent to each other.",
      "solution": "The only such $porcelain$ are $2/3, 3/2, (13 \\pm \\sqrt{601})/12$.\n\n\\textbf{First solution:}\nLet $butterflynet$ and $lanternlight$ be the curves\n$pumpkinlace = porcelain giraffewing^2 + porcelain giraffewing + \\frac{1}{24}$\nand $giraffewing = porcelain pumpkinlace^2 + porcelain pumpkinlace + \\frac{1}{24}$, respectively,\nand let $harvestmoon$ be the line $pumpkinlace = giraffewing$.\nWe consider three cases.\n\nIf $butterflynet$ is tangent to $harvestmoon$, then the point of tangency $(giraffewing,giraffewing)$ satisfies\n\\[\n2 porcelain giraffewing + porcelain = 1, \\qquad giraffewing = porcelain giraffewing^2 + porcelain giraffewing + \\frac{1}{24};\n\\]\nby symmetry, $lanternlight$ is tangent to $harvestmoon$ there, so $butterflynet$ and $lanternlight$ are tangent.\nWriting $porcelain = 1/(2giraffewing+1)$ in the first equation and substituting into\nthe second, we must have\n\\[\ngiraffewing = \\frac{giraffewing^2+giraffewing}{2giraffewing+1} + \\frac{1}{24},\n\\]\nwhich simplifies to $0 = 24giraffewing^2 - 2giraffewing - 1\n= (6giraffewing+1)(4giraffewing-1)$, or $giraffewing \\in \\{1/4, -1/6\\}$. This yields\n$porcelain = 1/(2giraffewing+1) \\in \\{2/3, 3/2\\}$.\n\nIf $butterflynet$ does not intersect $harvestmoon$, then $butterflynet$ and $lanternlight$ are separated by $harvestmoon$\nand so cannot be tangent.\n\nIf $butterflynet$ intersects $harvestmoon$ in two distinct points $whistlegrass, thimbleberry$, then it is not\ntangent to $harvestmoon$ at either point. Suppose at one of these points, say $whistlegrass$,\nthe tangent to $butterflynet$ is perpendicular to $harvestmoon$; then by symmetry, the same\nwill be true of $lanternlight$, so $butterflynet$ and $lanternlight$ will be tangent at $whistlegrass$. In this\ncase, the point $whistlegrass = (giraffewing,giraffewing)$ satisfies\n\\[\n2 porcelain giraffewing + porcelain = -1, \\qquad giraffewing = porcelain giraffewing^2 + porcelain giraffewing + \\frac{1}{24};\n\\]\nwriting $porcelain = -1/(2giraffewing+1)$ in the first equation and substituting into\nthe second, we have\n\\[\ngiraffewing = -\\frac{giraffewing^2+giraffewing}{2giraffewing+1} + \\frac{1}{24},\n\\]\nor\n$giraffewing = (-23 \\pm \\sqrt{601})/72$.\nThis yields\n$porcelain = -1/(2giraffewing+1) = (13 \\pm \\sqrt{601})/12$.\n\nIf instead the tangents to $butterflynet$ at $whistlegrass, thimbleberry$ are not perpendicular to\n$harvestmoon$, then we claim there cannot be any point where $butterflynet$ and $lanternlight$ are tangent.\nIndeed, if we count intersections of $butterflynet$ and $lanternlight$ (by using $butterflynet$ to\nsubstitute for $pumpkinlace$ in $lanternlight$, then solving for $pumpkinlace$), we get at most four\nsolutions counting multiplicity. Two of these are $whistlegrass$ and $thimbleberry$,\nand any point of tangency counts for two more. However, off of $harvestmoon$,\nany point of tangency would have a mirror image which is also a point of\ntangency, and there cannot be six solutions. Hence we have now found all\npossible $porcelain$.\n\n\\textbf{Second solution:}\nFor any nonzero value of $porcelain$, the two\nconics will intersect in four points in the complex projective plane\n$\\mathbb{P}^2(\\mathbb{C})$. To determine the\n$pumpkinlace$-coordinates of these intersection points, subtract the two equations\nto obtain\n\\[\n(pumpkinlace-giraffewing) = porcelain(giraffewing-pumpkinlace)(giraffewing+pumpkinlace) + porcelain(giraffewing-pumpkinlace).\n\\]\nTherefore, at\na point of intersection we have either $giraffewing=pumpkinlace$, or $giraffewing = -1/porcelain - (pumpkinlace+1)$.\nSubstituting these two possible linear conditions into the second\nequation shows that the $pumpkinlace$-coordinate of a point of intersection is a\nroot of either\n$cobblestone(pumpkinlace) = porcelain pumpkinlace^2+(porcelain-1)pumpkinlace + 1/24$ or\n$dandelionseed(pumpkinlace) = porcelain pumpkinlace^2 + (porcelain+1) pumpkinlace + 25/24 +1/porcelain$.\n\nIf two curves\nare tangent, then the $pumpkinlace$-coordinates of at least two of the\nintersection points will coincide; the converse is also true because one of the\ncurves is the graph of a function in $giraffewing$. The coincidence\noccurs precisely when either\nthe discriminant of at least one of $cobblestone$ or $dandelionseed$ is zero, or\nthere is a common root of $cobblestone$ and $dandelionseed$. Computing the discriminants of $cobblestone$\nand $dandelionseed$ yields (up to constant factors) $caterpillar(porcelain)=6porcelain^2 -\n13porcelain + 6$ and $dragonfly(porcelain)=6porcelain^2 - 13porcelain - 18$, respectively.\nIf on the other hand $cobblestone$ and $dandelionseed$ have a common root, it must\nbe also a root of $dandelionseed(pumpkinlace) - cobblestone(pumpkinlace) = 2pumpkinlace +1 + 1/porcelain$,\nyielding $pumpkinlace = -(1+porcelain)/(2porcelain)$ and\n$0 = cobblestone(pumpkinlace) = -dragonfly(porcelain)/(24 porcelain)$.\n\nThus the values of\n$porcelain$ for which the two curves are tangent must be contained in the\nset of zeros of $caterpillar$ and $dragonfly$, namely $2/3$, $3/2$, and\n$(13\\pm\\sqrt{601})/12$.\n\n\\textbf{Remark:}\nThe fact that the two conics in $\\mathbb{P}^2(\\CC)$ meet in four points, counted\nwith multiplicities, is a special case of \\emph{B\\'ezout's theorem}: two\ncurves in $\\mathbb{P}^2(\\CC)$ of degrees $storyteller, marshmallow$ and not sharing any common component\nmeet in exactly $storyteller marshmallow$ points when counted with multiplicity.\n\nMany solvers were surprised that the proposers chose the parameter $1/24$\nto give two rational roots and two nonrational roots. In fact, they had\nno choice in the matter: attempting to make all four roots rational\nby replacing $1/24$ by $blackpepper$ amounts\nto asking for $blackpepper^2 + blackpepper$ and $blackpepper^2 + blackpepper + 1$ to be perfect\nsquares. This cannot happen outside of trivial cases ($blackpepper = 0, -1$)\nultimately because the elliptic curve 24A1 (in Cremona's notation)\nover $\\mathbb{Q}$ has rank $0$. (Thanks to Noam Elkies for providing this\ncomputation.)\n\nHowever, there are choices that make the radical milder,\ne.g., $blackpepper = 1/3$  gives\n$blackpepper^2 + blackpepper = 4/9$ and $blackpepper^2 + blackpepper + 1 = 13/9$,\nwhile $blackpepper = 3/5$ gives $blackpepper^2 + blackpepper = 24/25$\nand $blackpepper^2 + blackpepper + 1 = 49/25$. "
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "nonxsymbol",
        "y": "nonysymbol",
        "L": "curvature",
        "C_1": "straightone",
        "C_2": "straighttwo",
        "P_1": "infiniteone",
        "P_2": "infinitetwo",
        "Q_1": "linearone",
        "Q_2": "lineartwo",
        "f_1": "constantone",
        "f_2": "constanttwo",
        "m": "largemvalue",
        "n": "largenvalue",
        "\\alpha": "lastvalue",
        "\\beta": "lastletter"
      },
      "question": "Find all values of $lastvalue$ for which the curves $nonysymbol = lastvalue nonxsymbol^2 +\nlastvalue nonxsymbol + \\frac{1}{24}$ and $nonxsymbol = lastvalue nonysymbol^2 + lastvalue nonysymbol + \\frac{1}{24}$\nare tangent to each other.",
      "solution": ""
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "L": "pouydfgh",
        "C_1": "xkqjweor",
        "C_2": "mvcnzlsa",
        "P_1": "rtugmnbv",
        "P_2": "wqzxclka",
        "Q_1": "lkjhasdf",
        "Q_2": "mnbvcxza",
        "f_1": "ghtrplkj",
        "f_2": "vbnhytre",
        "m": "zasxdrfv",
        "n": "plmoknij",
        "\\alpha": "qwertyui",
        "\\beta": "asdfghjk"
      },
      "question": "Find all values of $qwertyui$ for which the curves $hjgrksla = qwertyui qzxwvtnp^2 +\nqwertyui qzxwvtnp + \\frac{1}{24}$ and $qzxwvtnp = qwertyui hjgrksla^2 + qwertyui hjgrksla + \\frac{1}{24}$\nare tangent to each other.",
      "solution": "The only such $qwertyui$ are $2/3, 3/2, (13 \\pm \\sqrt{601})/12$.\n\n\\textbf{First solution:}\nLet $xkqjweor$ and $mvcnzlsa$ be the curves\n$hjgrksla=qwertyui qzxwvtnp^2 + qwertyui qzxwvtnp + \\frac{1}{24}$\nand $qzxwvtnp=qwertyui hjgrksla^2 + qwertyui hjgrksla + \\frac{1}{24}$, respectively,\nand let $pouydfgh$ be the line $hjgrksla=qzxwvtnp$.\nWe consider three cases.\n\nIf $xkqjweor$ is tangent to $pouydfgh$, then the point of tangency $(qzxwvtnp,qzxwvtnp)$ satisfies\n\\[\n2qwertyui qzxwvtnp + qwertyui = 1, \\qquad qzxwvtnp = qwertyui qzxwvtnp^2 + qwertyui qzxwvtnp + \\frac{1}{24};\n\\]\nby symmetry, $mvcnzlsa$ is tangent to $pouydfgh$ there, so $xkqjweor$ and $mvcnzlsa$ are tangent.\nWriting $qwertyui = 1/(2qzxwvtnp+1)$ in the first equation and substituting into\nthe second, we must have\n\\[\nqzxwvtnp = \\frac{qzxwvtnp^2+qzxwvtnp}{2qzxwvtnp+1} + \\frac{1}{24},\n\\]\nwhich simplifies to $0 = 24qzxwvtnp^2 - 2qzxwvtnp - 1\n= (6qzxwvtnp+1)(4qzxwvtnp-1)$, or $qzxwvtnp \\in \\{1/4, -1/6\\}$. This yields\n$qwertyui = 1/(2qzxwvtnp+1) \\in \\{2/3, 3/2\\}$.\n\nIf $xkqjweor$ does not intersect $pouydfgh$, then $xkqjweor$ and $mvcnzlsa$ are separated by $pouydfgh$\nand so cannot be tangent.\n\nIf $xkqjweor$ intersects $pouydfgh$ in two distinct points $rtugmnbv, wqzxclka$, then it is not\ntangent to $pouydfgh$ at either point. Suppose at one of these points, say $rtugmnbv$,\nthe tangent to $xkqjweor$ is perpendicular to $pouydfgh$; then by symmetry, the same\nwill be true of $mvcnzlsa$, so $xkqjweor$ and $mvcnzlsa$ will be tangent at $rtugmnbv$. In this\ncase, the point $rtugmnbv = (qzxwvtnp,qzxwvtnp)$ satisfies\n\\[\n2 qwertyui qzxwvtnp + qwertyui = -1, \\qquad qzxwvtnp = qwertyui qzxwvtnp^2 + qwertyui qzxwvtnp + \\frac{1}{24};\n\\]\nwriting $qwertyui = -1/(2qzxwvtnp+1)$ in the first equation and substituting into\nthe second, we have\n\\[\nqzxwvtnp = -\\frac{qzxwvtnp^2+qzxwvtnp}{2qzxwvtnp+1} + \\frac{1}{24},\n\\]\nor\n$qzxwvtnp = (-23 \\pm \\sqrt{601})/72$.\nThis yields\n$qwertyui = -1/(2qzxwvtnp+1) = (13 \\pm \\sqrt{601})/12$.\n\nIf instead the tangents to $xkqjweor$ at $rtugmnbv, wqzxclka$ are not perpendicular to\n$pouydfgh$, then we claim there cannot be any point where $xkqjweor$ and $mvcnzlsa$ are tangent.\nIndeed, if we count intersections of $xkqjweor$ and $mvcnzlsa$ (by using $xkqjweor$ to\nsubstitute for $hjgrksla$ in $mvcnzlsa$, then solving for $hjgrksla$), we get at most four\nsolutions counting multiplicity. Two of these are $rtugmnbv$ and $wqzxclka$,\nand any point of tangency counts for two more. However, off of $pouydfgh$,\nany point of tangency would have a mirror image which is also a point of\ntangency, and there cannot be six solutions. Hence we have now found all\npossible $qwertyui$.\n\n\\textbf{Second solution:}\nFor any nonzero value of $qwertyui$, the two\nconics will intersect in four points in the complex projective plane\n$\\mathbb{P}^2(\\mathbb{C})$. To determine the\n$hjgrksla$-coordinates of these intersection points, subtract the two equations\nto obtain\n\\[\n(hjgrksla-qzxwvtnp) = qwertyui(qzxwvtnp-hjgrksla)(qzxwvtnp+hjgrksla) + qwertyui(qzxwvtnp-hjgrksla).\n\\]\nTherefore, at\na point of intersection we have either $qzxwvtnp=hjgrksla$, or $qzxwvtnp = -1/qwertyui - (hjgrksla+1)$.\nSubstituting these two possible linear conditions into the second\nequation shows that the $hjgrksla$-coordinate of a point of intersection is a\nroot of either\n$lkjhasdf(hjgrksla) = qwertyui hjgrksla^2+(qwertyui-1)hjgrksla + 1/24$ or\n$mnbvcxza(hjgrksla) = qwertyui hjgrksla^2 + (qwertyui+1) hjgrksla + 25/24 +1/qwertyui$.\n\nIf two curves\nare tangent, then the $hjgrksla$-coordinates of at least two of the\nintersection points will coincide; the converse is also true because one of the\ncurves is the graph of a function in $qzxwvtnp$. The coincidence\noccurs precisely when either\nthe discriminant of at least one of $lkjhasdf$ or $mnbvcxza$ is zero, or\nthere is a common root of $lkjhasdf$ and $mnbvcxza$. Computing the discriminants of $lkjhasdf$\nand $mnbvcxza$ yields (up to constant factors) $ghtrplkj(qwertyui)=6qwertyui^2 -\n13qwertyui + 6$ and $vbnhytre(qwertyui)=6qwertyui^2 - 13qwertyui - 18$, respectively.\nIf on the other hand $lkjhasdf$ and $mnbvcxza$ have a common root, it must\nbe also a root of $mnbvcxza(hjgrksla) - lkjhasdf(hjgrksla) = 2hjgrksla +1 + 1/qwertyui$,\nyielding $hjgrksla = -(1+qwertyui)/(2qwertyui)$ and\n$0 = lkjhasdf(hjgrksla) = -vbnhytre(qwertyui)/(24 qwertyui)$.\n\nThus the values of\n$qwertyui$ for which the two curves are tangent must be contained in the\nset of zeros of $ghtrplkj$ and $vbnhytre$, namely $2/3$, $3/2$, and\n$(13\\pm\\sqrt{601})/12$.\n\n\\textbf{Remark:}\nThe fact that the two conics in $\\mathbb{P}^2(\\CC)$ meet in four points, counted\nwith multiplicities, is a special case of \\emph{B\\'ezout's theorem}: two\ncurves in $\\mathbb{P}^2(\\CC)$ of degrees $zasxdrfv, plmoknij$ and not sharing any common component\nmeet in exactly $zasxdrfv plmoknij$ points when counted with multiplicity.\n\nMany solvers were surprised that the proposers chose the parameter $1/24$\nto give two rational roots and two nonrational roots. In fact, they had\nno choice in the matter: attempting to make all four roots rational\nby replacing $1/24$ by $asdfghjk$ amounts\nto asking for $asdfghjk^2 + asdfghjk$ and $asdfghjk^2 + asdfghjk + 1$ to be perfect\nsquares. This cannot happen outside of trivial cases ($asdfghjk = 0, -1$)\nultimately because the elliptic curve 24A1 (in Cremona's notation)\nover $\\mathbb{Q}$ has rank $0$. (Thanks to Noam Elkies for providing this\ncomputation.)\n\nHowever, there are choices that make the radical milder,\ne.g., $asdfghjk = 1/3$  gives\n$asdfghjk^2 + asdfghjk = 4/9$ and $asdfghjk^2 + asdfghjk + 1 = 13/9$,\nwhile $asdfghjk = 3/5$ gives $asdfghjk^2 + asdfghjk = 24/25$\nand $asdfghjk^2 + asdfghjk + 1 = 49/25$. }"
    },
    "kernel_variant": {
      "question": "Determine all real numbers \\(\\alpha\\) for which the two curves\n\\[\n\\begin{cases}\nC_{1}:& y=\\alpha x^{2}+\\alpha x+\\dfrac{1}{24},\\\\[4pt]\nC_{2}:& x=\\alpha y^{2}+\\alpha y+\\dfrac{1}{24}\n\\end{cases}\n\\]\nare tangent to each other.",
      "solution": "We assume throughout that \\(\\alpha\\neq0\\); if \\(\\alpha=0\\) the two curves are the horizontal and vertical lines \\(y=1/24\\) and \\(x=1/24\\), which obviously are not tangent.\n\n------------------------------------------------------------\n1.  A point of tangency must lie on the diagonal \\(L:\\,y=x\\)\n------------------------------------------------------------\n\nWrite the two curves as the zero–sets of the polynomials\n\\[\n  f(x,y)=y-\\alpha x^{2}-\\alpha x-\\frac1{24},\\qquad\n  g(x,y)=x-\\alpha y^{2}-\\alpha y-\\frac1{24}.\n\\]\nAt a common point \\((x,y)\\) the curves are tangent precisely when their\ngradients are parallel, i.e.\n\\[\\nabla f\\times\\nabla g=\\bigl(-2\\alpha x-\\alpha,1\\bigr)\\times\\bigl(1,-2\\alpha y-\\alpha\\bigr)=0.\\]\nThe vanishing of the two–dimensional cross–product gives\n\\[\n(2\\alpha x+\\alpha)(2\\alpha y+\\alpha)=1.\\tag{1}\n\\]\nIntroduce\n\\[p=2x+1,\\qquad q=2y+1,\\qquad c=\\frac1{24}.\\]\nEquation (1) becomes\n\\[\\alpha^{2}pq=1.\\tag{2}\\]\nUsing \\(f=0\\) and \\(g=0\\) we can eliminate the constant term \\(c\\).  Subtracting the two relations gives\n\\[\nq-p=\\alpha\\frac{(p-1)^2-(q-1)^2}{2}+\\alpha(p-q)\n     =(p-q)\\,\\alpha\\Bigl(\\frac{p+q}{2}\\Bigr).\\tag{3}\n\\]\nSuppose, for contradiction, that \\(p\\ne q\\) (equivalently, \\(x\\ne y\\)).\nDividing (3) by \\(p-q\\) yields\n\\[-1=\\alpha\\frac{p+q}{2}\\quad\\Longrightarrow\\quad p+q=-\\frac2{\\alpha}.\\tag{4}\\]\nTogether with (2) this gives\n\\[(p-q)^2=(p+q)^2-4pq=\\frac{4}{\\alpha^{2}}-\\frac{4}{\\alpha^{2}}=0,\\]\na contradiction.  Hence \\(p=q\\), so \\(x=y\\).  Therefore every real\npoint at which \\(C_{1}\\) and \\(C_{2}\\) are tangent lies on the line\n\\(L:\\,y=x\\).\n\n-----------------------------------------------------------------\n2.  The slope of the common tangent\n-----------------------------------------------------------------\n\nBecause \\(C_{1}\\) and \\(C_{2}\\) are exchanged by the reflection \\((x,y)\\mapsto(y,x)\\), the common tangent at a point \\((t,t)\\in L\\) must be either parallel to \\(L\\) (slope \\(1\\)) or perpendicular to \\(L\\) (slope \\(-1\\)); otherwise the reflected tangent would be distinct, forcing two different tangent directions at the same point.\n\nFor \\(C_{1}\\) the derivative is\n\\[\\frac{dy}{dx}=2\\alpha x+\\alpha,\\]\nso at \\((t,t)\\) the slope is \\(2\\alpha t+\\alpha\\).\nWe therefore consider two cases.\n\n-----------------------------------------------------------------\nCase A: slope \\(1\\) (tangent parallel to \\(L\\))\n-----------------------------------------------------------------\n\\[2\\alpha t+\\alpha = 1 \\;\\Longrightarrow\\; \\alpha=\\frac{1}{2t+1}.\\tag{5}\\]\nBecause \\((t,t)\\) lies on \\(C_{1}\\),\n\\[t = \\alpha t^{2}+\\alpha t + \\frac1{24}.\\tag{6}\\]\nSubstituting (5) in (6) gives\n\\[t=\\frac{t^{2}+t}{2t+1}+\\frac1{24}\n  \\;\\Longrightarrow\\;24t^{2}-2t-1=0,\n\\]\nwhose solutions are\n\\[t\\in\\left\\{-\\frac16,\\,\\frac14\\right\\}.\\]\nUsing (5):\n\\[\n t=\\frac14\\;\\Rightarrow\\;\\alpha=\\frac23;\\qquad\n t=-\\frac16\\;\\Rightarrow\\;\\alpha=\\frac32.\n\\]\n\n-----------------------------------------------------------------\nCase B: slope \\(-1\\) (tangent perpendicular to \\(L\\))\n-----------------------------------------------------------------\n\\[2\\alpha t+\\alpha = -1 \\;\\Longrightarrow\\; \\alpha=-\\frac{1}{2t+1}.\\tag{7}\\]\nInsert (7) into (6):\n\\[t=-\\frac{t^{2}+t}{2t+1}+\\frac1{24}\n  \\;\\Longrightarrow\\;72t^{2}+46t-1=0.\n\\]\nHence\n\\[t=\\frac{-23\\pm\\sqrt{601}}{72},\\]\nwhich in turn gives, via (7),\n\\[\\alpha=\\frac{13\\pm\\sqrt{601}}{12}.\n\\]\n\n------------------------------------------------------------\n3.  No other possibilities\n------------------------------------------------------------\nThe preliminary argument in Section 1 shows that a tangency must occur on \\(L\\); Section 2 shows that on \\(L\\) only the two slopes \\(1\\) and \\(-1\\) are possible.  Consequently the four values of \\(\\alpha\\) found above exhaust all cases.\n\n------------------------------------------------------------\nAnswer\n------------------------------------------------------------\n\\[\\boxed{\\displaystyle \\alpha\\in\\Bigl\\{\\tfrac23,\\;\\tfrac32,\\;\\tfrac{13+\\sqrt{601}}{12},\\;\\tfrac{13-\\sqrt{601}}{12}\\Bigr\\}.}\\]",
      "_meta": {
        "core_steps": [
          "Exploit the x↔y symmetry; reduce the search for tangency to points on the line y = x or to symmetric off-diagonal pairs.",
          "At y = x impose slope-1 tangency:  derivative 2αx+α equals 1 and point lies on the curve; solve this system.",
          "For off-diagonal tangency use slope −1 (perpendicular to y = x); solve the analogous system.",
          "Invoke intersection-count/Bezout to argue that no further tangencies can exist once the above candidates are found."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "common additive constant in both quadratic equations",
            "original": "1/24"
          },
          "slot2": {
            "description": "the fact that the coefficients of x^2 (resp. y^2) and x (resp. y) are identical; any single parameter could replace α provided the same parameter appears in both powers",
            "original": "α appears as the coefficient of both the quadratic and the linear terms"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}