{
  "index": "1961-A-1",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "1. The graph of the equation \\( x^{y}=y^{x} \\) in the first quadrant (i.e., the region where \\( x>0 \\) and \\( y>0 \\) ) consists of a straight line and a curve. Find the coordinates of the intersection point of the line and the curve.",
  "solution": "Solution. In the first quadrant the given equation is equivalent to\n(1)\n\\[\n\\frac{1}{y} \\log y=\\frac{1}{x} \\log x .\n\\]\n\nConsider the function given by\n\\[\nf(t)=\\frac{1}{t} \\log t \\text { for } t>0\n\\]\n\nSince \\( f^{\\prime}(t)=(1-\\log t) / t^{2} \\), it is clear that \\( f \\) is strictly increasing for \\( t \\leq e \\), is strictly decreasing for \\( t \\geq e \\), and achieves its maximum value \\( e^{-1} \\) for \\( t \\) \\( =e \\). Moreover, \\( f(t) \\rightarrow-\\infty \\) as \\( t \\rightarrow 0 \\), and \\( f(t) \\rightarrow 0 \\) as \\( t \\rightarrow \\infty \\). It follows that for \\( \\alpha \\in\\left(0, e^{-1}\\right) \\) the equation \\( f(t)=\\alpha \\) has two solutions, one in \\( (1, e) \\), the other in ( \\( e, \\infty \\) ). For \\( \\alpha \\) near 0 , the lower solution is just above 1 and the upper solution is large. As \\( \\alpha \\) increases to \\( e^{-1} \\), the lower solution increases to \\( e \\) and the upper solution decreases to \\( e \\). Therefore, the locus (1) consists of the line \\( y=x \\) and a curve \\( M \\) lying in the quadrant \\( x>1, y>1 \\) and asymptotic to the line \\( x=1 \\) and \\( y=1 \\), as shown. \\( M \\) is evidently symmetric in the line \\( y=x \\) and crosses that line at \\( \\langle e, e\\rangle \\).\n\nWe can establish the smoothness of the curve \\( M \\) analytically. If \\( F \\) is a smooth function (say \\( C^{\\infty} \\) ) defined on an open set in \\( \\mathbf{R}^{2} \\) then the level sets (contour lines) of \\( F \\) are smooth curves except possibly at critical points of \\( F \\) (i.e., points where both partial derivatives of \\( F \\) vanish). At a critical point the structure of the level sets is the same as that of the level sets of the second Taylor polynomial of \\( F \\), provided the second degree terms are a non-degenerate quadratic form.\n\nLet\n\\[\nF=\\frac{1}{y} \\log y-\\frac{1}{x} \\log x \\text { for } x>0, y>0 .\n\\]\n\nThen\n\\[\nF_{1}^{\\prime}=\\frac{-1}{x^{2}}(1-\\log x) . \\quad F_{2}^{\\prime}=\\frac{1}{y^{2}}(1-\\log y)\n\\]\nand the only critical point of \\( F \\) is at \\( \\langle\\boldsymbol{e} . e\\rangle \\). At this point the second Taylor polynomial of \\( F \\) is\n\\[\n\\frac{1}{2 e^{3}}\\left((x-e)^{2}-(y-e)^{2}\\right) .\n\\]\n\nThis is a non-degenerate quadratic form that vanishes along the lines of slopes +1 and -1 through the point \\( \\langle\\rho, \\bullet\\rangle \\). It follows that the level sets of \\( F \\) are everywhere smooth curves except the one through 〈e. e〉 which is locally the union of two smooth curves having slopes +1 and -1 at that point. Since this level set is the required locus, this proves that the curve \\( M \\) described above is smooth.\n\nRemarks. The problem was discussed at length by E. J. Moulton, in \"The Real Function Defined by \\( x^{v}=y^{\\prime} \\). American Mathematical Monthly. vol. 23 (1916). pages 233-237. R. Robinson Rowe (Jourual of Recreational Mathematics. vol. 3 (1970). pages 176-178) calls the curve \\( M \\) the \"mutuabola.\"\n\nFor a discussion of level sets. see R. C. Buck. Advanced Calculus, McGraw-Hill. New York. 1965, page 349 ff.",
  "vars": [
    "x",
    "y",
    "t"
  ],
  "params": [
    "f",
    "F",
    "M",
    "\\\\alpha",
    "F_1",
    "F_2",
    "\\\\rho"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "t": "variablet",
        "f": "functionfsmall",
        "F": "functionfcapital",
        "M": "curvemutual",
        "\\alpha": "paramalpha",
        "F_1": "partialonef",
        "F_2": "partialtwof",
        "\\rho": "paramrho"
      },
      "question": "1. The graph of the equation \\( variablex^{variabley}=variabley^{variablex} \\) in the first quadrant (i.e., the region where \\( variablex>0 \\) and \\( variabley>0 \\) ) consists of a straight line and a curve. Find the coordinates of the intersection point of the line and the curve.",
      "solution": "Solution. In the first quadrant the given equation is equivalent to\n(1)\n\\[\n\\frac{1}{variabley} \\log variabley=\\frac{1}{variablex} \\log variablex .\n\\]\n\nConsider the function given by\n\\[\nfunctionfsmall(variablet)=\\frac{1}{variablet} \\log variablet \\text { for } variablet>0\n\\]\n\nSince \\( functionfsmall^{\\prime}(variablet)=(1-\\log variablet) / variablet^{2} \\), it is clear that \\( functionfsmall \\) is strictly increasing for \\( variablet \\leq e \\), is strictly decreasing for \\( variablet \\geq e \\), and achieves its maximum value \\( e^{-1} \\) for \\( variablet =e \\). Moreover, \\( functionfsmall(variablet) \\rightarrow-\\infty \\) as \\( variablet \\rightarrow 0 \\), and \\( functionfsmall(variablet) \\rightarrow 0 \\) as \\( variablet \\rightarrow \\infty \\). It follows that for \\( paramalpha \\in\\left(0, e^{-1}\\right) \\) the equation \\( functionfsmall(variablet)=paramalpha \\) has two solutions, one in \\( (1, e) \\), the other in \\( ( e, \\infty ) \\). For \\( paramalpha \\) near 0 , the lower solution is just above 1 and the upper solution is large. As \\( paramalpha \\) increases to \\( e^{-1} \\), the lower solution increases to \\( e \\) and the upper solution decreases to \\( e \\). Therefore, the locus (1) consists of the line \\( variabley=variablex \\) and a curve \\( curvemutual \\) lying in the quadrant \\( variablex>1, variabley>1 \\) and asymptotic to the line \\( variablex=1 \\) and \\( variabley=1 \\), as shown. \\( curvemutual \\) is evidently symmetric in the line \\( variabley=variablex \\) and crosses that line at \\( \\langle e, e\\rangle \\).\n\nWe can establish the smoothness of the curve \\( curvemutual \\) analytically. If \\( functionfcapital \\) is a smooth function (say \\( C^{\\infty} \\) ) defined on an open set in \\( \\mathbf{R}^{2} \\) then the level sets (contour lines) of \\( functionfcapital \\) are smooth curves except possibly at critical points of \\( functionfcapital \\) (i.e., points where both partial derivatives of \\( functionfcapital \\) vanish). At a critical point the structure of the level sets is the same as that of the level sets of the second Taylor polynomial of \\( functionfcapital \\), provided the second degree terms are a non-degenerate quadratic form.\n\nLet\n\\[\nfunctionfcapital=\\frac{1}{variabley} \\log variabley-\\frac{1}{variablex} \\log variablex \\text { for } variablex>0, variabley>0 .\n\\]\n\nThen\n\\[\npartialonef^{\\prime}=\\frac{-1}{variablex^{2}}(1-\\log variablex) . \\quad partialtwof^{\\prime}=\\frac{1}{variabley^{2}}(1-\\log variabley)\n\\]\nand the only critical point of \\( functionfcapital \\) is at \\( \\langle e , e\\rangle \\). At this point the second Taylor polynomial of \\( functionfcapital \\) is\n\\[\n\\frac{1}{2 e^{3}}\\left((variablex-e)^{2}-(variabley-e)^{2}\\right) .\n\\]\n\nThis is a non-degenerate quadratic form that vanishes along the lines of slopes +1 and -1 through the point \\( \\langle paramrho, \\bullet\\rangle \\). It follows that the level sets of \\( functionfcapital \\) are everywhere smooth curves except the one through 〈e, e〉 which is locally the union of two smooth curves having slopes +1 and -1 at that point. Since this level set is the required locus, this proves that the curve \\( curvemutual \\) described above is smooth.\n\nRemarks. The problem was discussed at length by E. J. Moulton, in \"The Real Function Defined by \\( variablex^{v}=variabley^{\\prime} \\). American Mathematical Monthly. vol. 23 (1916). pages 233-237. R. Robinson Rowe (Journal of Recreational Mathematics. vol. 3 (1970). pages 176-178) calls the curve \\( curvemutual \\) the \"mutuabola.\"\n\nFor a discussion of level sets, see R. C. Buck, Advanced Calculus, McGraw-Hill, New York, 1965, page 349 ff."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "umbrella",
        "y": "courtyard",
        "t": "pineapple",
        "f": "notebooke",
        "F": "playground",
        "M": "watermelon",
        "\\alpha": "lavenderia",
        "F_1": "playgroundone",
        "F_2": "playgroundtwo",
        "\\rho": "caterpillar"
      },
      "question": "1. The graph of the equation \\( umbrella^{courtyard}=courtyard^{umbrella} \\) in the first quadrant (i.e., the region where \\( umbrella>0 \\) and \\( courtyard>0 \\) ) consists of a straight line and a curve. Find the coordinates of the intersection point of the line and the curve.",
      "solution": "Solution. In the first quadrant the given equation is equivalent to\n(1)\n\\[\n\\frac{1}{courtyard} \\log courtyard=\\frac{1}{umbrella} \\log umbrella .\n\\]\n\nConsider the function given by\n\\[\nnotebooke(pineapple)=\\frac{1}{pineapple} \\log pineapple \\text { for } pineapple>0\n\\]\n\nSince \\( notebooke^{\\prime}(pineapple)=(1-\\log pineapple) / pineapple^{2} \\), it is clear that \\( notebooke \\) is strictly increasing for \\( pineapple \\leq e \\), is strictly decreasing for \\( pineapple \\geq e \\), and achieves its maximum value \\( e^{-1} \\) for \\( pineapple=e \\). Moreover, \\( notebooke(pineapple) \\rightarrow-\\infty \\) as \\( pineapple \\rightarrow 0 \\), and \\( notebooke(pineapple) \\rightarrow 0 \\) as \\( pineapple \\rightarrow \\infty \\). It follows that for \\( lavenderia \\in\\left(0, e^{-1}\\right) \\) the equation \\( notebooke(pineapple)=lavenderia \\) has two solutions, one in \\( (1, e) \\), the other in \\( ( e, \\infty ) \\). For \\( lavenderia \\) near 0, the lower solution is just above 1 and the upper solution is large. As \\( lavenderia \\) increases to \\( e^{-1} \\), the lower solution increases to \\( e \\) and the upper solution decreases to \\( e \\). Therefore, the locus (1) consists of the line \\( courtyard=umbrella \\) and a curve \\( watermelon \\) lying in the quadrant \\( umbrella>1, courtyard>1 \\) and asymptotic to the line \\( umbrella=1 \\) and \\( courtyard=1 \\), as shown. \\( watermelon \\) is evidently symmetric in the line \\( courtyard=umbrella \\) and crosses that line at \\( \\langle e, e\\rangle \\).\n\nWe can establish the smoothness of the curve \\( watermelon \\) analytically. If \\( playground \\) is a smooth function (say \\( C^{\\infty} \\) ) defined on an open set in \\( \\mathbf{R}^{2} \\) then the level sets (contour lines) of \\( playground \\) are smooth curves except possibly at critical points of \\( playground \\) (i.e., points where both partial derivatives of \\( playground \\) vanish). At a critical point the structure of the level sets is the same as that of the level sets of the second Taylor polynomial of \\( playground \\), provided the second degree terms are a non-degenerate quadratic form.\n\nLet\n\\[\nplayground=\\frac{1}{courtyard} \\log courtyard-\\frac{1}{umbrella} \\log umbrella \\text { for } umbrella>0, courtyard>0 .\n\\]\n\nThen\n\\[\nplaygroundone^{\\prime}=\\frac{-1}{umbrella^{2}}(1-\\log umbrella) . \\quad playgroundtwo^{\\prime}=\\frac{1}{courtyard^{2}}(1-\\log courtyard)\n\\]\nand the only critical point of \\( playground \\) is at \\( \\langle\\boldsymbol{e} . e\\rangle \\). At this point the second Taylor polynomial of \\( playground \\) is\n\\[\n\\frac{1}{2 e^{3}}\\left((umbrella-e)^{2}-(courtyard-e)^{2}\\right) .\n\\]\n\nThis is a non-degenerate quadratic form that vanishes along the lines of slopes +1 and -1 through the point \\( \\langle caterpillar, \\bullet\\rangle \\). It follows that the level sets of \\( playground \\) are everywhere smooth curves except the one through 〈e. e〉 which is locally the union of two smooth curves having slopes +1 and -1 at that point. Since this level set is the required locus, this proves that the curve \\( watermelon \\) described above is smooth.\n\nRemarks. The problem was discussed at length by E. J. Moulton, in \"The Real Function Defined by \\( umbrella^{v}=courtyard^{\\prime} \\). American Mathematical Monthly. vol. 23 (1916). pages 233-237. R. Robinson Rowe (Jourual of Recreational Mathematics. vol. 3 (1970). pages 176-178) calls the curve \\( watermelon \\) the \"mutuabola.\"\n\nFor a discussion of level sets, see R. C. Buck. Advanced Calculus, McGraw-Hill. New York. 1965, page 349 ff."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knownvalue",
        "y": "constantval",
        "t": "staticparam",
        "f": "nonmapping",
        "F": "nonrelation",
        "M": "straightpath",
        "\\alpha": "fixedconstant",
        "F_1": "nonrelationa",
        "F_2": "nonrelationb",
        "\\rho": "emptiness"
      },
      "question": "1. The graph of the equation \\( knownvalue^{constantval}=constantval^{knownvalue} \\) in the first quadrant (i.e., the region where \\( knownvalue>0 \\) and \\( constantval>0 \\) ) consists of a straight line and a curve. Find the coordinates of the intersection point of the line and the curve.",
      "solution": "Solution. In the first quadrant the given equation is equivalent to\n(1)\n\\[\n\\frac{1}{constantval} \\log constantval=\\frac{1}{knownvalue} \\log knownvalue .\n\\]\n\nConsider the function given by\n\\[\nnonmapping(staticparam)=\\frac{1}{staticparam} \\log staticparam \\text { for } staticparam>0\n\\]\n\nSince \\( nonmapping^{\\prime}(staticparam)=(1-\\log staticparam) / staticparam^{2} \\), it is clear that \\( nonmapping \\) is strictly increasing for \\( staticparam \\leq e \\), is strictly decreasing for \\( staticparam \\geq e \\), and achieves its maximum value \\( e^{-1} \\) for \\( staticparam =e \\). Moreover, \\( nonmapping(staticparam) \\rightarrow-\\infty \\) as \\( staticparam \\rightarrow 0 \\), and \\( nonmapping(staticparam) \\rightarrow 0 \\) as \\( staticparam \\rightarrow \\infty \\). It follows that for \\( fixedconstant \\in\\left(0, e^{-1}\\right) \\) the equation \\( nonmapping(staticparam)=fixedconstant \\) has two solutions, one in \\( (1, e) \\), the other in \\( ( e, \\infty ) \\). For \\( fixedconstant \\) near 0 , the lower solution is just above 1 and the upper solution is large. As \\( fixedconstant \\) increases to \\( e^{-1} \\), the lower solution increases to \\( e \\) and the upper solution decreases to \\( e \\). Therefore, the locus (1) consists of the line \\( constantval=knownvalue \\) and a curve \\( straightpath \\) lying in the quadrant \\( knownvalue>1, constantval>1 \\) and asymptotic to the line \\( knownvalue=1 \\) and \\( constantval=1 \\), as shown. \\( straightpath \\) is evidently symmetric in the line \\( constantval=knownvalue \\) and crosses that line at \\( \\langle e, e\\rangle \\).\n\nWe can establish the smoothness of the curve \\( straightpath \\) analytically. If \\( nonrelation \\) is a smooth function (say \\( C^{\\infty} \\) ) defined on an open set in \\( \\mathbf{R}^{2} \\) then the level sets (contour lines) of \\( nonrelation \\) are smooth curves except possibly at critical points of \\( nonrelation \\) (i.e., points where both partial derivatives of \\( nonrelation \\) vanish). At a critical point the structure of the level sets is the same as that of the level sets of the second Taylor polynomial of \\( nonrelation \\), provided the second degree terms are a non-degenerate quadratic form.\n\nLet\n\\[\nnonrelation=\\frac{1}{constantval} \\log constantval-\\frac{1}{knownvalue} \\log knownvalue \\text { for } knownvalue>0, constantval>0 .\n\\]\n\nThen\n\\[\nnonrelationa^{\\prime}=\\frac{-1}{knownvalue^{2}}(1-\\log knownvalue) . \\quad nonrelationb^{\\prime}=\\frac{1}{constantval^{2}}(1-\\log constantval)\n\\]\nand the only critical point of \\( nonrelation \\) is at \\( \\langle\\boldsymbol{e} . e\\rangle \\). At this point the second Taylor polynomial of \\( nonrelation \\) is\n\\[\n\\frac{1}{2 e^{3}}\\left((knownvalue-e)^{2}-(constantval-e)^{2}\\right) .\n\\]\n\nThis is a non-degenerate quadratic form that vanishes along the lines of slopes +1 and -1 through the point \\( \\langle emptiness, \\bullet\\rangle \\). It follows that the level sets of \\( nonrelation \\) are everywhere smooth curves except the one through 〈e. e〉 which is locally the union of two smooth curves having slopes +1 and -1 at that point. Since this level set is the required locus, this proves that the curve \\( straightpath \\) described above is smooth.\n\nRemarks. The problem was discussed at length by E. J. Moulton, in \"The Real Function Defined by \\( knownvalue^{v}=constantval^{\\prime} \\). American Mathematical Monthly. vol. 23 (1916). pages 233-237. R. Robinson Rowe (Jourual of Recreational Mathematics. vol. 3 (1970). pages 176-178) calls the curve \\( straightpath \\) the \"mutuabola.\"\n\nFor a discussion of level sets. see R. C. Buck. Advanced Calculus, McGraw-Hill. New York. 1965, page 349 ff."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "t": "mfldkjeu",
        "f": "prywndqc",
        "F": "uvcsgmlk",
        "M": "bzrthxpa",
        "\\alpha": "\\qnrldkfe",
        "F_1": "lmskqjtr",
        "F_2": "zpwvqany",
        "\\rho": "\\rcpvyzfn"
      },
      "question": "1. The graph of the equation \\( qzxwvtnp^{hjgrksla}=hjgrksla^{qzxwvtnp} \\) in the first quadrant (i.e., the region where \\( qzxwvtnp>0 \\) and \\( hjgrksla>0 \\) ) consists of a straight line and a curve. Find the coordinates of the intersection point of the line and the curve.",
      "solution": "Solution. In the first quadrant the given equation is equivalent to\n(1)\n\\[\n\\frac{1}{hjgrksla} \\log hjgrksla=\\frac{1}{qzxwvtnp} \\log qzxwvtnp .\n\\]\n\nConsider the function given by\n\\[\nprywndqc(mfldkjeu)=\\frac{1}{mfldkjeu} \\log mfldkjeu \\text { for } mfldkjeu>0\n\\]\n\nSince \\( prywndqc^{\\prime}(mfldkjeu)=(1-\\log mfldkjeu) / mfldkjeu^{2} \\), it is clear that \\( prywndqc \\) is strictly increasing for \\( mfldkjeu \\leq e \\), is strictly decreasing for \\( mfldkjeu \\geq e \\), and achieves its maximum value \\( e^{-1} \\) for \\( mfldkjeu =e \\). Moreover, \\( prywndqc(mfldkjeu) \\rightarrow-\\infty \\) as \\( mfldkjeu \\rightarrow 0 \\), and \\( prywndqc(mfldkjeu) \\rightarrow 0 \\) as \\( mfldkjeu \\rightarrow \\infty \\). It follows that for \\( \\qnrldkfe \\in\\left(0, e^{-1}\\right) \\) the equation \\( prywndqc(mfldkjeu)=\\qnrldkfe \\) has two solutions, one in \\( (1, e) \\), the other in ( \\( e, \\infty \\) ). For \\( \\qnrldkfe \\) near 0 , the lower solution is just above 1 and the upper solution is large. As \\( \\qnrldkfe \\) increases to \\( e^{-1} \\), the lower solution increases to \\( e \\) and the upper solution decreases to \\( e \\). Therefore, the locus (1) consists of the line \\( hjgrksla=qzxwvtnp \\) and a curve \\( bzrthxpa \\) lying in the quadrant \\( qzxwvtnp>1, hjgrksla>1 \\) and asymptotic to the line \\( qzxwvtnp=1 \\) and \\( hjgrksla=1 \\), as shown. \\( bzrthxpa \\) is evidently symmetric in the line \\( hjgrksla=qzxwvtnp \\) and crosses that line at \\( \\langle e, e\\rangle \\).\n\nWe can establish the smoothness of the curve \\( bzrthxpa \\) analytically. If \\( uvcsgmlk \\) is a smooth function (say \\( C^{\\infty} \\) ) defined on an open set in \\( \\mathbf{R}^{2} \\) then the level sets (contour lines) of \\( uvcsgmlk \\) are smooth curves except possibly at critical points of \\( uvcsgmlk \\) (i.e., points where both partial derivatives of \\( uvcsgmlk \\) vanish). At a critical point the structure of the level sets is the same as that of the level sets of the second Taylor polynomial of \\( uvcsgmlk \\), provided the second degree terms are a non-degenerate quadratic form.\n\nLet\n\\[\nuvcsgmlk=\\frac{1}{hjgrksla} \\log hjgrksla-\\frac{1}{qzxwvtnp} \\log qzxwvtnp \\text { for } qzxwvtnp>0, hjgrksla>0 .\n\\]\n\nThen\n\\[\nlmskqjtr^{\\prime}=\\frac{-1}{qzxwvtnp^{2}}(1-\\log qzxwvtnp) . \\quad zpwvqany^{\\prime}=\\frac{1}{hjgrksla^{2}}(1-\\log hjgrksla)\n\\]\nand the only critical point of \\( uvcsgmlk \\) is at \\( \\langle e, e\\rangle \\). At this point the second Taylor polynomial of \\( uvcsgmlk \\) is\n\\[\n\\frac{1}{2 e^{3}}\\left((qzxwvtnp-e)^{2}-(hjgrksla-e)^{2}\\right) .\n\\]\n\nThis is a non-degenerate quadratic form that vanishes along the lines of slopes +1 and -1 through the point \\( \\langle\\rcpvyzfn, \\bullet\\rangle \\). It follows that the level sets of \\( uvcsgmlk \\) are everywhere smooth curves except the one through 〈e. e〉 which is locally the union of two smooth curves having slopes +1 and -1 at that point. Since this level set is the required locus, this proves that the curve \\( bzrthxpa \\) described above is smooth.\n\nRemarks. The problem was discussed at length by E. J. Moulton, in \"The Real Function Defined by \\( qzxwvtnp^{hjgrksla}=hjgrksla^{qzxwvtnp} \\). American Mathematical Monthly. vol. 23 (1916). pages 233-237. R. Robinson Rowe (Jourual of Recreational Mathematics. vol. 3 (1970). pages 176-178) calls the curve \\( bzrthxpa \\) the \"mutuabola.\"\n\nFor a discussion of level sets. see R. C. Buck. Advanced Calculus, McGraw-Hill. New York. 1965, page 349 ff."
    },
    "kernel_variant": {
      "question": "For positive real numbers (u,v,w) with u,v,w>0 consider the system  \n\n        u^{v}=v^{w}=w^{u}.                                (★)  \n\n(a)  Show that every solution of (★) either lies on the main diagonal u=v=w or belongs to a second, distinct three–variable branch.  \n(b)  Prove that this second branch is a C¹-smooth two–dimensional surface throughout the open first octant, including its single point of contact with the diagonal.  \n(c)  Locate that unique common point and determine the tangent plane to the surface there.\n\n────────────────────────────────────────────────────────",
      "solution": "Throughout put  \n\n        f(t)=\\dfrac{\\ln t}{t},  t>0.                                 (1)\n\nStep 1.  Functional reformulation.  \nTaking natural logarithms of (★) is permissible because all variables are positive.  We obtain  \n\n        v\\,\\ln u=u\\,\\ln v,  w\\,\\ln v=v\\,\\ln w.                     (2)\n\nDividing respectively by u v>0 and v w>0 gives  \n\n        f(u)=f(v)=f(w).                                              (3)\n\nHence any solution is a triple on which f attains the same value three times.\n\nStep 2.  Monotonicity of f.  \nCompute  \n\n        f'(t)=\\dfrac{1-\\ln t}{t^{2}}.                                (4)\n\nNote that f'>0 on (0,e), f'(e)=0, and f'<0 on (e,∞).  Thus f is strictly increasing up to t=e and strictly decreasing thereafter, reaching its sole maximum f(e)=1/e at t=e.  Observe that f is one-to-one on each interval I₁:=(0,e] and I₂:=[e,∞).\n\nStep 3.  Classification of solutions.  \nBecause f assumes each value at most once on I₁ and at most once on I₂, equality (3) forces two possibilities.\n\n(a) All three variables lie in the same monotone region I₁ or I₂.  Since f is injective there, we must have u=v=w.  These points form the diagonal branch.  \n\n(b) At least one variable is from I₁ and at least one from I₂.  In that case not all three can coincide, yielding a second branch S that is symmetric under any permutation of (u,v,w).  Since every coordinate exceeds 1 (because f(t)<0 when t<1, while the common value in (3) is non-negative), S actually lies in the sub-octant u>1, v>1, w>1.\n\nStep 4.  Smoothness away from t=e.  \nIntroduce the mapping  \n\n        F(u,v,w)=(f(u)-f(v), f(v)-f(w))    :  (0,∞)³→ℝ².            (5)\n\nOn S we have F=0.  Compute the Jacobian with respect to (v,w):\n\n        J= ∂(F₁,F₂)/∂(v,w)=\n        [ -f'(v)    0\n           1       -f'(w) ].                                   (6)\n\nWhenever (v,w)≠(e,e) at least one derivative f' is non-zero, so det J=f'(v)f'(w)≠0.  By the implicit-function theorem S is C^∞ in a neighbourhood of every point for which (v,w)≠(e,e); equivalently, wherever not all coordinates equal e.\n\nStep 5.  Behaviour at the joining point.  \nWe now identify the point common to both branches and check differentiability there.\n\n5.1 Coordinates.  \nIf a point belongs to both branches we require simultaneously u=v=w (diagonal) and at least one coordinate in I₁ and one in I₂ (second branch).  The only way both statements can hold is to sit exactly at the boundary t=e.  It follows that the unique common point is  \n\n        (u,v,w)=(e,e,e).                                            (7)\n\n5.2 Tangent geometry at (e,e,e).  \nSet u=e+α, v=e+β, w=e+γ with small α,β,γ.  Using the quadratic Taylor expansion  \n\n        f(e+δ)=\\frac{1}{e}-\\frac{δ^{2}}{2e^{3}}+O(δ^{3}),          (8)\n\nand substituting into (3) we obtain  \n\n        α²=β²+O(3), β²=γ²+O(3)                                     (9)\n\nwhere O(3) abbreviates cubic terms in (α,β,γ).  Hence  \n\n        β=±α+o(α), γ=±β+o(β).                                      (10)\n\nBecause S is characterised by the presence of coordinates on opposite sides of e, the admissible choice is to take an odd number of minus signs; without loss of generality pick β=−α+o(α), γ=−β+o(α)=α+o(α).  Eliminating little-o terms we derive the linear relation  \n\n        α+β=0                                                       (11)\n\ntogether with α−γ=0.  Translating back to (u,v,w) this gives the tangent plane at (e,e,e):\n\n        (u−e)+(v−e)=0 and (u−e)−(w−e)=0 ⇒ u=e, v+w=2e.         (12)\n\nHence S meets the diagonal transversally, and the first derivatives of the local parametrisation exist; consequently S is C¹ at (e,e,e).\n\nStep 6.  Summary.  \nThe solution set of u^{v}=v^{w}=w^{u} in the positive octant is the disjoint union of  \n • the straight line u=v=w>0, and  \n • a C¹-smooth surface S lying in u>1, v>1, w>1, symmetric under permutations, which touches the diagonal only at (e,e,e) with tangent plane given by (12).  This completes the proof.\n\n────────────────────────────────────────────────────────",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.067604",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}