{
  "index": "1959-B-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "5. Find the equation of the smallest sphere which is tangent to both of the lines: (i) \\( x=t+1, y=2 t+4, z=-3 t+5 \\), and (ii) \\( x=4 t-12, y= \\) \\( -t+8, z=t+17 \\).",
  "solution": "First Solution. Let the given lines be \\( l \\) and \\( m \\). Then there is a unique segment \\( P Q \\) perpendicular to both lines with \\( P \\) on \\( l \\) and \\( Q \\) on \\( m \\). The required sphere has \\( P Q \\) as its diameter.\n\nSuppose \\( l \\) and \\( m \\) are given in terms of a parameter \\( t \\) by a \\( +t \\mathrm{v} \\) and \\( \\mathbf{b}+\\boldsymbol{t} \\mathbf{w} \\), respectively, where \\( \\mathbf{a}, \\mathbf{b}, \\mathbf{v} \\), and \\( \\mathbf{w} \\) are vectors. If the lines are not parallel, \\( \\mathbf{v} \\) and \\( \\mathbf{w} \\) are linearly independent. Since \\( P Q \\) is perpendicular to both lines, it has the direction of \\( \\mathbf{v} \\times \\mathbf{w} \\), say \\( \\overrightarrow{P Q}=\\rho \\mathbf{v} \\times \\mathbf{w} \\), where \\( \\rho \\) is a scalar. Let \\( P \\) and \\( Q \\) be the points \\( \\mathbf{a}+\\sigma \\mathbf{v} \\) and \\( \\mathbf{b}+\\tau \\mathbf{w} \\), respectively. Then \\( \\overrightarrow{P Q}=\\mathbf{b}-\\mathbf{a}-\\sigma \\mathbf{v}+\\tau \\mathbf{w} \\) and\n\\[\n\\mathbf{a}-\\mathbf{b}=-\\rho(\\mathbf{v} \\times \\mathbf{w})-\\sigma \\mathbf{v}+\\tau \\mathbf{w}\n\\]\n\nHence we can calculate \\( \\rho, \\sigma \\), and \\( \\tau \\) by expressing \\( \\mathbf{a}-\\mathbf{b} \\) in terms of the independent vectors \\( \\mathbf{v} \\times \\mathbf{w}, \\mathbf{v} \\) and \\( \\mathbf{w} \\). Then the center of the required sphere is at\n\\[\n\\mathbf{a}+\\sigma \\mathbf{v}+\\frac{1}{2} \\rho(\\mathbf{v} \\times \\mathbf{w})\n\\]\nand its radius is\n\\[\n\\frac{1}{2}|\\rho|\\|\\mathbf{v} \\times w\\| .\n\\]\n\nFor the example in question, \\( \\mathbf{a}=\\langle 1,4,5\\rangle \\mathbf{b}=\\langle-12,8,17\\rangle \\), \\( \\mathbf{v}=\\langle 1,2,-3\\rangle, \\mathbf{w}=\\langle 4,-1,1\\rangle \\) and \\( \\mathbf{v} \\times \\mathbf{w}=\\langle-1,-13,-9\\rangle \\). Then \\( \\rho, \\sigma \\) and \\( \\tau \\) are found from the equations\n\\[\n\\begin{aligned}\n13 & =\\rho-\\sigma+4 \\tau \\\\\n-4 & =13 \\rho-2 \\sigma-\\tau \\\\\n-12 & =9 \\rho+3 \\sigma+\\tau\n\\end{aligned}\n\\]\nwhich give\n\\[\n\\rho=\\frac{-147}{251}, \\quad \\sigma=\\frac{-782}{251}, \\quad \\tau=\\frac{657}{251} .\n\\]\n\nThe center of the sphere is therefore at\n\\[\n\\begin{array}{l}\n\\langle 1,4,5\\rangle-\\frac{782}{251}\\langle 1,2,-3\\rangle-\\frac{147}{502}\\langle-1,-13,-9\\rangle \\\\\n=\\frac{1}{502}\\langle-915,791,8525\\rangle .\n\\end{array}\n\\]\n\nThe square of the radius is\n\\[\n\\frac{1}{4} \\rho^{2}\\|\\overrightarrow{\\mathbf{v}} \\times \\overrightarrow{\\mathbf{w}}\\|^{2}=\\left(\\frac{147}{502}\\right)^{2}(251)=\\frac{147^{2}}{1004}\n\\]\n\nThe equation of the sphere is\n\\[\n(502 x+915)^{2}+(502 y-791)^{2}+(502 z-8525)^{2}=251(147)^{2}\n\\]\n\nSecond Solution. Let \\( P \\) and \\( Q \\) be chosen on \\( l \\) and \\( m \\), respectively, so that \\( P Q \\) is as short as possible. Then \\( P Q \\) is perpendicular to each of the lines \\( l \\) and \\( m \\). The sphere with \\( P Q \\) as diameter is the required sphere, for it is tangent to \\( l \\) and \\( m \\) and no smaller sphere intersects both \\( l \\) and \\( m \\).\n\nIf \\( P \\) is the point \\( \\langle s+1,2 s+4,-3 s+5\\rangle \\) and \\( Q \\) is the point \\( \\langle 4 t-12,-t+8, t+17\\rangle \\), then\n\\[\n\\begin{aligned}\n|P Q|^{2} & =(s-4 t+13)^{2}+(2 s+t-4)^{2}+(-3 s-t-12)^{2} \\\\\n& =14 s^{2}+2 s t+18 t^{2}+82 s-88 t+329\n\\end{aligned}\n\\]\n\nThe quadratic terms of this function are positive definite, so the minimum is achieved at the unique point at which both partial derivatives vanish. Hence the equations\n\\[\n\\begin{array}{l}\n28 s+2 t+82=0 \\\\\n2 s+36 t-88=0\n\\end{array}\n\\]\ndetermine the desired values of \\( s \\) and \\( t \\), which turn out to be \\( -782 / 251 \\) and \\( 657 / 251 \\), respectively.\n\nWe can now find \\( P \\) and \\( Q \\) and continue as in the first solution:\nRemark. The complicated arithmetic was not intended by the examination committee. The second equation of (i) was intended to be \\( y= \\) \\( 2 t-4 \\). Then in the first solution, we find \\( \\rho=-1, \\sigma=-2, \\tau=+3 \\). In the second solution the parameters turn out to be \\( s=-2 \\) and \\( t=3 \\).",
  "vars": [
    "x",
    "y",
    "z",
    "t",
    "s"
  ],
  "params": [
    "l",
    "m",
    "P",
    "Q",
    "a",
    "b",
    "v",
    "w",
    "\\\\rho",
    "\\\\sigma",
    "\\\\tau"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "axisxcoor",
        "y": "axisycoor",
        "z": "axiszcoor",
        "t": "paramtvar",
        "s": "paramsvar",
        "l": "linefirst",
        "m": "linesecond",
        "P": "pointpnt",
        "Q": "pointqnt",
        "a": "vectoraorg",
        "b": "vectorborg",
        "v": "vectorvdir",
        "w": "vectorwdir",
        "\\rho": "scalerho",
        "\\sigma": "scalesigma",
        "\\tau": "scaletau"
      },
      "question": "5. Find the equation of the smallest sphere which is tangent to both of the lines: (i) \\( axisxcoor=paramtvar+1, axisycoor=2 paramtvar+4, axiszcoor=-3 paramtvar+5 \\), and (ii) \\( axisxcoor=4 paramtvar-12, axisycoor= \\) \\( -paramtvar+8, axiszcoor=paramtvar+17 \\).",
      "solution": "First Solution. Let the given lines be \\( linefirst \\) and \\( linesecond \\). Then there is a unique segment \\( pointpnt pointqnt \\) perpendicular to both lines with \\( pointpnt \\) on \\( linefirst \\) and \\( pointqnt \\) on \\( linesecond \\). The required sphere has \\( pointpnt pointqnt \\) as its diameter.\n\nSuppose \\( linefirst \\) and \\( linesecond \\) are given in terms of a parameter \\( paramtvar \\) by \\( vectoraorg +paramtvar \\mathrm{vectorvdir} \\) and \\( \\mathbf{vectorborg}+\\boldsymbol{paramtvar} \\mathbf{vectorwdir} \\), respectively, where \\( \\mathbf{vectoraorg}, \\mathbf{vectorborg}, \\mathbf{vectorvdir} \\), and \\( \\mathbf{vectorwdir} \\) are vectors. If the lines are not parallel, \\( \\mathbf{vectorvdir} \\) and \\( \\mathbf{vectorwdir} \\) are linearly independent. Since \\( pointpnt pointqnt \\) is perpendicular to both lines, it has the direction of \\( \\mathbf{vectorvdir} \\times \\mathbf{vectorwdir} \\), say \\( \\overrightarrow{pointpnt pointqnt}=scalerho \\mathbf{vectorvdir} \\times \\mathbf{vectorwdir} \\), where \\( scalerho \\) is a scalar. Let \\( pointpnt \\) and \\( pointqnt \\) be the points \\( \\mathbf{vectoraorg}+scalesigma \\mathbf{vectorvdir} \\) and \\( \\mathbf{vectorborg}+scaletau \\mathbf{vectorwdir} \\), respectively. Then \\( \\overrightarrow{pointpnt pointqnt}=\\mathbf{vectorborg}-\\mathbf{vectoraorg}-scalesigma \\mathbf{vectorvdir}+scaletau \\mathbf{vectorwdir} \\) and\n\\[\n\\mathbf{vectoraorg}-\\mathbf{vectorborg}=-scalerho(\\mathbf{vectorvdir} \\times \\mathbf{vectorwdir})-scalesigma \\mathbf{vectorvdir}+scaletau \\mathbf{vectorwdir}\n\\]\n\nHence we can calculate \\( scalerho, scalesigma \\), and \\( scaletau \\) by expressing \\( \\mathbf{vectoraorg}-\\mathbf{vectorborg} \\) in terms of the independent vectors \\( \\mathbf{vectorvdir} \\times \\mathbf{vectorwdir}, \\mathbf{vectorvdir} \\) and \\( \\mathbf{vectorwdir} \\). Then the center of the required sphere is at\n\\[\n\\mathbf{vectoraorg}+scalesigma \\mathbf{vectorvdir}+\\frac{1}{2} scalerho(\\mathbf{vectorvdir} \\times \\mathbf{vectorwdir})\n\\]\nand its radius is\n\\[\n\\frac{1}{2}|scalerho|\\|\\mathbf{vectorvdir} \\times vectorwdir\\| .\n\\]\n\nFor the example in question, \\( \\mathbf{vectoraorg}=\\langle 1,4,5\\rangle \\mathbf{vectorborg}=\\langle-12,8,17\\rangle \\), \\( \\mathbf{vectorvdir}=\\langle 1,2,-3\\rangle, \\mathbf{vectorwdir}=\\langle 4,-1,1\\rangle \\) and \\( \\mathbf{vectorvdir} \\times \\mathbf{vectorwdir}=\\langle-1,-13,-9\\rangle \\). Then \\( scalerho, scalesigma \\) and \\( scaletau \\) are found from the equations\n\\[\n\\begin{aligned}\n13 & =scalerho-scalesigma+4 scaletau \\\\\n-4 & =13 scalerho-2 scalesigma-scaletau \\\\\n-12 & =9 scalerho+3 scalesigma+scaletau\n\\end{aligned}\n\\]\nwhich give\n\\[\nscalerho=\\frac{-147}{251}, \\quad scalesigma=\\frac{-782}{251}, \\quad scaletau=\\frac{657}{251} .\n\\]\n\nThe center of the sphere is therefore at\n\\[\n\\begin{array}{l}\n\\langle 1,4,5\\rangle-\\frac{782}{251}\\langle 1,2,-3\\rangle-\\frac{147}{502}\\langle-1,-13,-9\\rangle \\\\\n=\\frac{1}{502}\\langle-915,791,8525\\rangle .\n\\end{array}\n\\]\n\nThe square of the radius is\n\\[\n\\frac{1}{4} scalerho^{2}\\|\\overrightarrow{\\mathbf{vectorvdir}} \\times \\overrightarrow{\\mathbf{vectorwdir}}\\|^{2}=\\left(\\frac{147}{502}\\right)^{2}(251)=\\frac{147^{2}}{1004}\n\\]\n\nThe equation of the sphere is\n\\[\n(502 axisxcoor+915)^{2}+(502 axisycoor-791)^{2}+(502 axiszcoor-8525)^{2}=251(147)^{2}\n\\]\n\nSecond Solution. Let \\( pointpnt \\) and \\( pointqnt \\) be chosen on \\( linefirst \\) and \\( linesecond \\), respectively, so that \\( pointpnt pointqnt \\) is as short as possible. Then \\( pointpnt pointqnt \\) is perpendicular to each of the lines \\( linefirst \\) and \\( linesecond \\). The sphere with \\( pointpnt pointqnt \\) as diameter is the required sphere, for it is tangent to \\( linefirst \\) and \\( linesecond \\) and no smaller sphere intersects both \\( linefirst \\) and \\( linesecond \\).\n\nIf \\( pointpnt \\) is the point \\( \\langle paramsvar+1,2 paramsvar+4,-3 paramsvar+5\\rangle \\) and \\( pointqnt \\) is the point \\( \\langle 4 paramtvar-12,-paramtvar+8, paramtvar+17\\rangle \\), then\n\\[\n\\begin{aligned}\n|pointpnt pointqnt|^{2} & =(paramsvar-4 paramtvar+13)^{2}+(2 paramsvar+paramtvar-4)^{2}+(-3 paramsvar-paramtvar-12)^{2} \\\\\n& =14 paramsvar^{2}+2 paramsvar paramtvar+18 paramtvar^{2}+82 paramsvar-88 paramtvar+329\n\\end{aligned}\n\\]\n\nThe quadratic terms of this function are positive definite, so the minimum is achieved at the unique point at which both partial derivatives vanish. Hence the equations\n\\[\n\\begin{array}{l}\n28 paramsvar+2 paramtvar+82=0 \\\\\n2 paramsvar+36 paramtvar-88=0\n\\end{array}\n\\]\ndetermine the desired values of \\( paramsvar \\) and \\( paramtvar \\), which turn out to be \\( -782 / 251 \\) and \\( 657 / 251 \\), respectively.\n\nWe can now find \\( pointpnt \\) and \\( pointqnt \\) and continue as in the first solution:\n\nRemark. The complicated arithmetic was not intended by the examination committee. The second equation of (i) was intended to be \\( axisycoor= \\) \\( 2 paramtvar-4 \\). Then in the first solution, we find \\( scalerho=-1, scalesigma=-2, scaletau=+3 \\). In the second solution the parameters turn out to be \\( paramsvar=-2 \\) and \\( paramtvar=3 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "y": "helicopter",
        "z": "marshmallow",
        "t": "strawberries",
        "s": "watercress",
        "l": "peppermint",
        "m": "butterscotch",
        "P": "blueberry",
        "Q": "cinnamon",
        "a": "adventure",
        "b": "kangaroo",
        "v": "lighthouse",
        "w": "sandcastle",
        "\\rho": "seashells",
        "\\sigma": "hummingbird",
        "\\tau": "jellybeans"
      },
      "question": "5. Find the equation of the smallest sphere which is tangent to both of the lines: (i) \\( pineapple=strawberries+1, helicopter=2 strawberries+4, marshmallow=-3 strawberries+5 \\), and (ii) \\( pineapple=4 strawberries-12, helicopter= \\) \\( -strawberries+8, marshmallow=strawberries+17 \\).",
      "solution": "First Solution. Let the given lines be \\( peppermint \\) and \\( butterscotch \\). Then there is a unique segment \\( blueberry cinnamon \\) perpendicular to both lines with \\( blueberry \\) on \\( peppermint \\) and \\( cinnamon \\) on \\( butterscotch \\). The required sphere has \\( blueberry cinnamon \\) as its diameter.\n\nSuppose \\( peppermint \\) and \\( butterscotch \\) are given in terms of a parameter \\( strawberries \\) by a \\( +strawberries \\mathrm{lighthouse} \\) and \\( \\mathbf{kangaroo}+\\boldsymbol{strawberries} \\mathbf{sandcastle} \\), respectively, where \\( \\mathbf{adventure}, \\mathbf{kangaroo}, \\mathbf{lighthouse} \\), and \\( \\mathbf{sandcastle} \\) are vectors. If the lines are not parallel, \\( \\mathbf{lighthouse} \\) and \\( \\mathbf{sandcastle} \\) are linearly independent. Since \\( blueberry cinnamon \\) is perpendicular to both lines, it has the direction of \\( \\mathbf{lighthouse} \\times \\mathbf{sandcastle} \\), say \\( \\overrightarrow{blueberry cinnamon}=seashells \\mathbf{lighthouse} \\times \\mathbf{sandcastle} \\), where \\( seashells \\) is a scalar. Let \\( blueberry \\) and \\( cinnamon \\) be the points \\( \\mathbf{adventure}+hummingbird \\mathbf{lighthouse} \\) and \\( \\mathbf{kangaroo}+jellybeans \\mathbf{sandcastle} \\), respectively. Then \\( \\overrightarrow{blueberry cinnamon}=\\mathbf{kangaroo}-\\mathbf{adventure}-hummingbird \\mathbf{lighthouse}+jellybeans \\mathbf{sandcastle} \\) and\n\\[\n\\mathbf{adventure}-\\mathbf{kangaroo}=-seashells(\\mathbf{lighthouse} \\times \\mathbf{sandcastle})-hummingbird \\mathbf{lighthouse}+jellybeans \\mathbf{sandcastle}\n\\]\n\nHence we can calculate \\( seashells, hummingbird \\), and \\( jellybeans \\) by expressing \\( \\mathbf{adventure}-\\mathbf{kangaroo} \\) in terms of the independent vectors \\( \\mathbf{lighthouse} \\times \\mathbf{sandcastle}, \\mathbf{lighthouse} \\) and \\( \\mathbf{sandcastle} \\). Then the center of the required sphere is at\n\\[\n\\mathbf{adventure}+hummingbird \\mathbf{lighthouse}+\\frac{1}{2} seashells(\\mathbf{lighthouse} \\times \\mathbf{sandcastle})\n\\]\nand its radius is\n\\[\n\\frac{1}{2}|seashells|\\|\\mathbf{lighthouse} \\times sandcastle\\| .\n\\]\n\nFor the example in question, \\( \\mathbf{adventure}=\\langle 1,4,5\\rangle \\mathbf{kangaroo}=\\langle-12,8,17\\rangle \\), \\( \\mathbf{lighthouse}=\\langle 1,2,-3\\rangle, \\mathbf{sandcastle}=\\langle 4,-1,1\\rangle \\) and \\( \\mathbf{lighthouse} \\times \\mathbf{sandcastle}=\\langle-1,-13,-9\\rangle \\). Then \\( seashells, hummingbird \\) and \\( jellybeans \\) are found from the equations\n\\[\n\\begin{aligned}\n13 &=seashells-hummingbird+4 jellybeans \\\\\n-4 &=13 seashells-2 hummingbird-jellybeans \\\\\n-12 &=9 seashells+3 hummingbird+jellybeans\n\\end{aligned}\n\\]\nwhich give\n\\[\nseashells=\\frac{-147}{251}, \\quad hummingbird=\\frac{-782}{251}, \\quad jellybeans=\\frac{657}{251} .\n\\]\n\nThe center of the sphere is therefore at\n\\[\n\\begin{array}{l}\n\\langle 1,4,5\\rangle-\\frac{782}{251}\\langle 1,2,-3\\rangle-\\frac{147}{502}\\langle-1,-13,-9\\rangle \\\\\n=\\frac{1}{502}\\langle-915,791,8525\\rangle .\n\\end{array}\n\\]\n\nThe square of the radius is\n\\[\n\\frac{1}{4} seashells^{2}\\|\\overrightarrow{\\mathbf{lighthouse}} \\times \\overrightarrow{\\mathbf{sandcastle}}\\|^{2}=\\left(\\frac{147}{502}\\right)^{2}(251)=\\frac{147^{2}}{1004}\n\\]\n\nThe equation of the sphere is\n\\[\n(502 pineapple+915)^{2}+(502 helicopter-791)^{2}+(502 marshmallow-8525)^{2}=251(147)^{2}\n\\]\n\nSecond Solution. Let \\( blueberry \\) and \\( cinnamon \\) be chosen on \\( peppermint \\) and \\( butterscotch \\), respectively, so that \\( blueberry cinnamon \\) is as short as possible. Then \\( blueberry cinnamon \\) is perpendicular to each of the lines \\( peppermint \\) and \\( butterscotch \\). The sphere with \\( blueberry cinnamon \\) as diameter is the required sphere, for it is tangent to \\( peppermint \\) and \\( butterscotch \\) and no smaller sphere intersects both \\( peppermint \\) and \\( butterscotch \\).\n\nIf \\( blueberry \\) is the point \\( \\langle watercress+1,2 watercress+4,-3 watercress+5\\rangle \\) and \\( cinnamon \\) is the point \\( \\langle 4 strawberries-12,-strawberries+8, strawberries+17\\rangle \\), then\n\\[\n\\begin{aligned}\n|blueberry cinnamon|^{2} & =(watercress-4 strawberries+13)^{2}+(2 watercress+strawberries-4)^{2}+(-3 watercress-strawberries-12)^{2} \\\\\n& =14 watercress^{2}+2 watercress strawberries+18 strawberries^{2}+82 watercress-88 strawberries+329\n\\end{aligned}\n\\]\n\nThe quadratic terms of this function are positive definite, so the minimum is achieved at the unique point at which both partial derivatives vanish. Hence the equations\n\\[\n\\begin{array}{l}\n28 watercress+2 strawberries+82=0 \\\\\n2 watercress+36 strawberries-88=0\n\\end{array}\n\\]\ndetermine the desired values of \\( watercress \\) and \\( strawberries \\), which turn out to be \\( -782 / 251 \\) and \\( 657 / 251 \\), respectively.\n\nWe can now find \\( blueberry \\) and \\( cinnamon \\) and continue as in the first solution:\nRemark. The complicated arithmetic was not intended by the examination committee. The second equation of (i) was intended to be \\( helicopter= \\) \\( 2 strawberries-4 \\). Then in the first solution, we find \\( seashells=-1, hummingbird=-2, jellybeans=+3 \\). In the second solution the parameters turn out to be \\( watercress=-2 \\) and \\( strawberries=3 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "certainty",
        "y": "definiteness",
        "z": "planarity",
        "t": "stillness",
        "s": "constancy",
        "l": "curvepath",
        "m": "circlepath",
        "P": "broadzone",
        "Q": "vastzone",
        "a": "endingpoint",
        "b": "terminalpt",
        "v": "stillvector",
        "w": "restvector",
        "\\rho": "\\vectorvalue",
        "\\sigma": "\\chaoslevel",
        "\\tau": "\\staticrate"
      },
      "question": "5. Find the equation of the smallest sphere which is tangent to both of the lines: (i) \\( certainty=stillness+1, definiteness=2\\, stillness+4, planarity=-3\\, stillness+5 \\), and (ii) \\( certainty=4\\, stillness-12, definiteness= -\\,stillness+8, planarity=stillness+17 \\).",
      "solution": "First Solution. Let the given lines be \\( curvepath \\) and \\( circlepath \\). Then there is a unique segment \\( broadzone \\, vastzone \\) perpendicular to both lines with \\( broadzone \\) on \\( curvepath \\) and \\( vastzone \\) on \\( circlepath \\). The required sphere has \\( broadzone \\, vastzone \\) as its diameter.\n\nSuppose \\( curvepath \\) and \\( circlepath \\) are given in terms of a parameter \\( stillness \\) by \\( \\mathbf{endingpoint}+stillness \\mathbf{stillvector} \\) and \\( \\mathbf{terminalpt}+stillness \\mathbf{restvector} \\), respectively, where \\( \\mathbf{endingpoint}, \\mathbf{terminalpt}, \\mathbf{stillvector} \\), and \\( \\mathbf{restvector} \\) are vectors. If the lines are not parallel, \\( \\mathbf{stillvector} \\) and \\( \\mathbf{restvector} \\) are linearly independent. Since \\( broadzone \\, vastzone \\) is perpendicular to both lines, it has the direction of \\( \\mathbf{stillvector} \\times \\mathbf{restvector} \\), say \\( \\overrightarrow{broadzone\\, vastzone}= \\vectorvalue \\, \\mathbf{stillvector} \\times \\mathbf{restvector} \\), where \\( \\vectorvalue \\) is a scalar. Let \\( broadzone \\) and \\( vastzone \\) be the points \\( \\mathbf{endingpoint}+ \\chaoslevel \\mathbf{stillvector} \\) and \\( \\mathbf{terminalpt}+ \\staticrate \\mathbf{restvector} \\), respectively. Then \\( \\overrightarrow{broadzone\\, vastzone}= \\mathbf{terminalpt}-\\mathbf{endingpoint}- \\chaoslevel \\mathbf{stillvector}+ \\staticrate \\mathbf{restvector} \\) and\n\\[\n\\mathbf{endingpoint}-\\mathbf{terminalpt}= -\\vectorvalue(\\mathbf{stillvector} \\times \\mathbf{restvector})-\\chaoslevel \\mathbf{stillvector}+ \\staticrate \\mathbf{restvector}\n\\]\n\nHence we can calculate \\( \\vectorvalue, \\chaoslevel \\), and \\( \\staticrate \\) by expressing \\( \\mathbf{endingpoint}-\\mathbf{terminalpt} \\) in terms of the independent vectors \\( \\mathbf{stillvector} \\times \\mathbf{restvector}, \\mathbf{stillvector} \\) and \\( \\mathbf{restvector} \\). Then the center of the required sphere is at\n\\[\n\\mathbf{endingpoint}+ \\chaoslevel \\mathbf{stillvector}+ \\frac{1}{2} \\vectorvalue(\\mathbf{stillvector} \\times \\mathbf{restvector})\n\\]\nand its radius is\n\\[\n\\frac{1}{2}|\\vectorvalue|\\|\\mathbf{stillvector} \\times \\mathbf{restvector}\\| .\n\\]\n\nFor the example in question, \\( \\mathbf{endingpoint}=\\langle 1,4,5\\rangle, \\mathbf{terminalpt}=\\langle-12,8,17\\rangle \\), \\( \\mathbf{stillvector}=\\langle 1,2,-3\\rangle, \\mathbf{restvector}=\\langle 4,-1,1\\rangle \\) and \\( \\mathbf{stillvector} \\times \\mathbf{restvector}=\\langle-1,-13,-9\\rangle \\). Then \\( \\vectorvalue, \\chaoslevel \\) and \\( \\staticrate \\) are found from the equations\n\\[\n\\begin{aligned}\n13 & =\\vectorvalue-\\chaoslevel+4 \\staticrate \\\\\n-4 & =13 \\vectorvalue-2 \\chaoslevel-\\staticrate \\\\\n-12 & =9 \\vectorvalue+3 \\chaoslevel+\\staticrate\n\\end{aligned}\n\\]\nwhich give\n\\[\\vectorvalue=\\frac{-147}{251}, \\quad \\chaoslevel=\\frac{-782}{251}, \\quad \\staticrate=\\frac{657}{251} .\\]\n\nThe center of the sphere is therefore at\n\\[\n\\begin{array}{l}\n\\langle 1,4,5\\rangle-\\frac{782}{251}\\langle 1,2,-3\\rangle-\\frac{147}{502}\\langle-1,-13,-9\\rangle \\\\\n=\\frac{1}{502}\\langle-915,791,8525\\rangle .\n\\end{array}\n\\]\n\nThe square of the radius is\n\\[\n\\frac{1}{4} \\vectorvalue^{2}\\|\\mathbf{stillvector} \\times \\mathbf{restvector}\\|^{2}=\\left(\\frac{147}{502}\\right)^{2}(251)=\\frac{147^{2}}{1004}\n\\]\n\nThe equation of the sphere is\n\\[\n(502 certainty+915)^{2}+(502 definiteness-791)^{2}+(502 planarity-8525)^{2}=251(147)^{2}\n\\]\n\nSecond Solution. Let \\( broadzone \\) and \\( vastzone \\) be chosen on \\( curvepath \\) and \\( circlepath \\), respectively, so that \\( broadzone \\, vastzone \\) is as short as possible. Then \\( broadzone \\, vastzone \\) is perpendicular to each of the lines \\( curvepath \\) and \\( circlepath \\). The sphere with \\( broadzone \\, vastzone \\) as diameter is the required sphere, for it is tangent to \\( curvepath \\) and \\( circlepath \\) and no smaller sphere intersects both \\( curvepath \\) and \\( circlepath \\).\n\nIf \\( broadzone \\) is the point \\( \\langle constancy+1,2\\, constancy+4,-3\\, constancy+5\\rangle \\) and \\( vastzone \\) is the point \\( \\langle 4\\, stillness-12,-stillness+8, stillness+17\\rangle \\), then\n\\[\n\\begin{aligned}\n|broadzone \\, vastzone|^{2} & =(constancy-4 \\, stillness+13)^{2}+(2 \\, constancy+stillness-4)^{2}+(-3 \\, constancy-stillness-12)^{2} \\\\\n& =14 \\, constancy^{2}+2\\, constancy\\, stillness+18\\, stillness^{2}+82\\, constancy-88\\, stillness+329\n\\end{aligned}\n\\]\n\nThe quadratic terms of this function are positive definite, so the minimum is achieved at the unique point at which both partial derivatives vanish. Hence the equations\n\\[\n\\begin{array}{l}\n28 \\, constancy+2\\, stillness+82=0 \\\\\n2\\, constancy+36\\, stillness-88=0\n\\end{array}\n\\]\ndetermine the desired values of \\( constancy \\) and \\( stillness \\), which turn out to be \\( -782 / 251 \\) and \\( 657 / 251 \\), respectively.\n\nWe can now find \\( broadzone \\) and \\( vastzone \\) and continue as in the first solution:\nRemark. The complicated arithmetic was not intended by the examination committee. The second equation of (i) was intended to be \\( definiteness=2\\, stillness-4 \\). Then in the first solution, we find \\( \\vectorvalue=-1, \\chaoslevel=-2, \\staticrate=+3 \\). In the second solution the parameters turn out to be \\( constancy=-2 \\) and \\( stillness=3 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "hzxvbqrm",
        "y": "glfstzpc",
        "z": "qrmhgxld",
        "t": "vmkclprj",
        "s": "npthwzga",
        "l": "bqrfmnlz",
        "m": "prstvkjh",
        "P": "jkdhwqzr",
        "Q": "fzmrkshv",
        "a": "grzgctsb",
        "b": "nljmqvch",
        "v": "plknsrjx",
        "w": "hdqtmfzc",
        "\\rho": "qzxwvtnp",
        "\\sigma": "hjgrksla",
        "\\tau": "vbxcpdwr"
      },
      "question": "5. Find the equation of the smallest sphere which is tangent to both of the lines: (i) \\( hzxvbqrm=vmkclprj+1, glfstzpc=2 vmkclprj+4, qrmhgxld=-3 vmkclprj+5 \\), and (ii) \\( hzxvbqrm=4 vmkclprj-12, glfstzpc= \\) \\( -vmkclprj+8, qrmhgxld=vmkclprj+17 \\).",
      "solution": "First Solution. Let the given lines be \\( bqrfmnlz \\) and \\( prstvkjh \\). Then there is a unique segment \\( jkdhwqzr fzmrkshv \\) perpendicular to both lines with \\( jkdhwqzr \\) on \\( bqrfmnlz \\) and \\( fzmrkshv \\) on \\( prstvkjh \\). The required sphere has \\( jkdhwqzr fzmrkshv \\) as its diameter.\n\nSuppose \\( bqrfmnlz \\) and \\( prstvkjh \\) are given in terms of a parameter \\( vmkclprj \\) by \\( \\mathbf{grzgctsb}+vmkclprj \\mathbf{plknsrjx} \\) and \\( \\mathbf{nljmqvch}+vmkclprj \\mathbf{hdqtmfzc} \\), respectively, where \\( \\mathbf{grzgctsb}, \\mathbf{nljmqvch}, \\mathbf{plknsrjx} \\), and \\( \\mathbf{hdqtmfzc} \\) are vectors. If the lines are not parallel, \\( \\mathbf{plknsrjx} \\) and \\( \\mathbf{hdqtmfzc} \\) are linearly independent. Since \\( jkdhwqzr fzmrkshv \\) is perpendicular to both lines, it has the direction of \\( \\mathbf{plknsrjx} \\times \\mathbf{hdqtmfzc} \\), say \\( \\overrightarrow{jkdhwqzr fzmrkshv}=qzxwvtnp \\mathbf{plknsrjx} \\times \\mathbf{hdqtmfzc} \\), where \\( qzxwvtnp \\) is a scalar. Let \\( jkdhwqzr \\) and \\( fzmrkshv \\) be the points \\( \\mathbf{grzgctsb}+hjgrksla \\mathbf{plknsrjx} \\) and \\( \\mathbf{nljmqvch}+vbxcpdwr \\mathbf{hdqtmfzc} \\), respectively. Then \\( \\overrightarrow{jkdhwqzr fzmrkshv}=\\mathbf{nljmqvch}-\\mathbf{grzgctsb}-hjgrksla \\mathbf{plknsrjx}+vbxcpdwr \\mathbf{hdqtmfzc} \\) and\n\\[\n\\mathbf{grzgctsb}-\\mathbf{nljmqvch}=-qzxwvtnp(\\mathbf{plknsrjx} \\times \\mathbf{hdqtmfzc})-hjgrksla \\mathbf{plknsrjx}+vbxcpdwr \\mathbf{hdqtmfzc}\n\\]\n\nHence we can calculate \\( qzxwvtnp, hjgrksla \\), and \\( vbxcpdwr \\) by expressing \\( \\mathbf{grzgctsb}-\\mathbf{nljmqvch} \\) in terms of the independent vectors \\( \\mathbf{plknsrjx} \\times \\mathbf{hdqtmfzc}, \\mathbf{plknsrjx} \\) and \\( \\mathbf{hdqtmfzc} \\). Then the center of the required sphere is at\n\\[\n\\mathbf{grzgctsb}+hjgrksla \\mathbf{plknsrjx}+\\frac{1}{2} qzxwvtnp(\\mathbf{plknsrjx} \\times \\mathbf{hdqtmfzc})\n\\]\nand its radius is\n\\[\n\\frac{1}{2}|qzxwvtnp|\\|\\mathbf{plknsrjx} \\times hdqtmfzc\\| .\n\\]\n\nFor the example in question, \\( \\mathbf{grzgctsb}=\\langle 1,4,5\\rangle \\mathbf{nljmqvch}=\\langle-12,8,17\\rangle \\), \\( \\mathbf{plknsrjx}=\\langle 1,2,-3\\rangle, \\mathbf{hdqtmfzc}=\\langle 4,-1,1\\rangle \\) and \\( \\mathbf{plknsrjx} \\times \\mathbf{hdqtmfzc}=\\langle-1,-13,-9\\rangle \\). Then \\( qzxwvtnp, hjgrksla \\) and \\( vbxcpdwr \\) are found from the equations\n\\[\n\\begin{aligned}\n13 & =qzxwvtnp-hjgrksla+4 vbxcpdwr \\\\\n-4 & =13 qzxwvtnp-2 hjgrksla-vbxcpdwr \\\\\n-12 & =9 qzxwvtnp+3 hjgrksla+vbxcpdwr\n\\end{aligned}\n\\]\nwhich give\n\\[\nqzxwvtnp=\\frac{-147}{251}, \\quad hjgrksla=\\frac{-782}{251}, \\quad vbxcpdwr=\\frac{657}{251} .\n\\]\n\nThe center of the sphere is therefore at\n\\[\n\\begin{array}{l}\n\\langle 1,4,5\\rangle-\\frac{782}{251}\\langle 1,2,-3\\rangle-\\frac{147}{502}\\langle-1,-13,-9\\rangle \\\\\n=\\frac{1}{502}\\langle-915,791,8525\\rangle .\n\\end{array}\n\\]\n\nThe square of the radius is\n\\[\n\\frac{1}{4} qzxwvtnp^{2}\\|\\overrightarrow{\\mathbf{plknsrjx}} \\times \\overrightarrow{\\mathbf{hdqtmfzc}}\\|^{2}=\\left(\\frac{147}{502}\\right)^{2}(251)=\\frac{147^{2}}{1004}\n\\]\n\nThe equation of the sphere is\n\\[\n(502 hzxvbqrm+915)^{2}+(502 glfstzpc-791)^{2}+(502 qrmhgxld-8525)^{2}=251(147)^{2}\n\\]\n\nSecond Solution. Let \\( jkdhwqzr \\) and \\( fzmrkshv \\) be chosen on \\( bqrfmnlz \\) and \\( prstvkjh \\), respectively, so that \\( jkdhwqzr fzmrkshv \\) is as short as possible. Then \\( jkdhwqzr fzmrkshv \\) is perpendicular to each of the lines \\( bqrfmnlz \\) and \\( prstvkjh \\). The sphere with \\( jkdhwqzr fzmrkshv \\) as diameter is the required sphere, for it is tangent to \\( bqrfmnlz \\) and \\( prstvkjh \\) and no smaller sphere intersects both \\( bqrfmnlz \\) and \\( prstvkjh \\).\n\nIf \\( jkdhwqzr \\) is the point \\( \\langle npthwzga+1,2 npthwzga+4,-3 npthwzga+5\\rangle \\) and \\( fzmrkshv \\) is the point \\( \\langle 4 vmkclprj-12,-vmkclprj+8, vmkclprj+17\\rangle \\), then\n\\[\n\\begin{aligned}\n|jkdhwqzr fzmrkshv|^{2} & =(npthwzga-4 vmkclprj+13)^{2}+(2 npthwzga+vmkclprj-4)^{2}+(-3 npthwzga-vmkclprj-12)^{2} \\\\\n& =14 npthwzga^{2}+2 npthwzga vmkclprj+18 vmkclprj^{2}+82 npthwzga-88 vmkclprj+329\n\\end{aligned}\n\\]\n\nThe quadratic terms of this function are positive definite, so the minimum is achieved at the unique point at which both partial derivatives vanish. Hence the equations\n\\[\n\\begin{array}{l}\n28 npthwzga+2 vmkclprj+82=0 \\\\\n2 npthwzga+36 vmkclprj-88=0\n\\end{array}\n\\]\ndetermine the desired values of \\( npthwzga \\) and \\( vmkclprj \\), which turn out to be \\( -782 / 251 \\) and \\( 657 / 251 \\), respectively.\n\nWe can now find \\( jkdhwqzr \\) and \\( fzmrkshv \\) and continue as in the first solution:\nRemark. The complicated arithmetic was not intended by the examination committee. The second equation of (i) was intended to be \\( glfstzpc= \\) \\( 2 vmkclprj-4 \\). Then in the first solution, we find \\( qzxwvtnp=-1, hjgrksla=-2, vbxcpdwr=+3 \\). In the second solution the parameters turn out to be \\( npthwzga=-2 \\) and \\( vmkclprj=3 \\)."
    },
    "kernel_variant": {
      "question": "In Euclidean four–space $\\mathbb R^{4}$ consider the three pairwise skew affine lines  \n\\[\n\\ell_{1}:\\;(x,y,z,w)=\\bigl(t,\\;1,\\;0,\\;0\\bigr),\\qquad t\\in\\mathbb R,\n\\]\n\\[\n\\ell_{2}:\\;(x,y,z,w)=\\bigl(0,\\;s,\\;1,\\;0\\bigr),\\qquad s\\in\\mathbb R,\n\\]\n\\[\n\\ell_{3}:\\;(x,y,z,w)=\\bigl(1,\\;0,\\;u,\\;0\\bigr),\\qquad u\\in\\mathbb R .\n\\]\n\nAmong all $3$–spheres in $\\mathbb R^{4}$ that are tangent to {\\em each} of the three lines, determine the unique sphere of smallest possible radius.  \nGive its equation in the standard form\n\\[\nS:\\;(x-a)^{2}+(y-b)^{2}+(z-c)^{2}+(w-d)^{2}=r^{2},\n\\]\nand state the centre $(a,b,c,d)$ as well as the radius $r$ in exact radical form.",
      "solution": "{\\bf Step 1.  Squared distances from a general point to the three lines.}  \nFor a point $C=(a,b,c,d)\\in\\mathbb R^{4}$ orthogonal projection onto the lines gives  \n\\[\n\\begin{aligned}\nd_{1}^{2}&=\\operatorname{dist}^{2}(C,\\ell_{1})\n        =(b-1)^{2}+c^{2}+d^{2},\\\\[4pt]\nd_{2}^{2}&=\\operatorname{dist}^{2}(C,\\ell_{2})\n        =a^{2}+(c-1)^{2}+d^{2},\\\\[4pt]\nd_{3}^{2}&=\\operatorname{dist}^{2}(C,\\ell_{3})\n        =(a-1)^{2}+b^{2}+d^{2}.\n\\end{aligned}\n\\]\nIf a $3$–sphere centred at $C$ is tangent simultaneously to $\\ell_{1},\\ell_{2},\\ell_{3}$ we must have\n\\[\nd_{1}^{2}=d_{2}^{2}=d_{3}^{2}=r^{2}.\n\\tag{1}\n\\]\n\n\\medskip\n{\\bf Step 2.  Algebraic form of the tangency constraints.}  \nEquality of the first two expressions in (1) yields\n\\[\n(b-1)^{2}+c^{2}=a^{2}+(c-1)^{2}\n\\;\\Longrightarrow\\;\ng_{1}(a,b,c):=(b-1)^{2}-a^{2}+2c-1=0.\n\\tag{2}\n\\]\nEquality of the first and the third gives\n\\[\n(b-1)^{2}+c^{2}=(a-1)^{2}+b^{2}\n\\;\\Longrightarrow\\;\ng_{2}(a,b,c):=(b-1)^{2}-a^{2}-b^{2}+c^{2}+2a-1=0.\n\\tag{3}\n\\]\nThe coordinate $d$ is unconstrained by (2)–(3).\n\n\\medskip\n{\\bf Step 3.  A symmetry–averaging argument implies $a=b=c$.}  \n\nLet $G=\\langle\\sigma\\rangle\\cong\\mathbb Z_{3}$ act on $\\mathbb R^{4}$ by cyclic permutation  \n\\[\n\\sigma:(x,y,z,w)\\longmapsto(y,z,x,w).\n\\]\nThe unordered set $\\{\\ell_{1},\\ell_{2},\\ell_{3}\\}$ is $G$–invariant.  Hence if a centre $C=(a,b,c,d)$ is feasible, then so are the two points $\\sigma C=(b,c,a,d)$ and $\\sigma^{2}C=(c,a,b,d)$, and all three share the {\\em same} radius $r$ by (1).  \n\nSet \n\\[\nf(b,c):=(b-1)^{2}+c^{2}\\qquad\\bigl(\\text{so }d_{1}^{2}=f(b,c)+d^{2}\\bigr).\n\\]\nThe function $f$ is a strictly convex quadratic form in the pair $(b,c)$; indeed its Hessian is $\\operatorname{diag}(2,2)\\succ 0$.  \nNow average the three orbit points:\n\\[\n\\overline C=\\frac13\\bigl(C+\\sigma C+\\sigma^{2}C\\bigr)\n           =\\Bigl(\\tfrac{a+b+c}{3},\\tfrac{a+b+c}{3},\\tfrac{a+b+c}{3},d\\Bigr).\n\\]\nBecause $d$ is unchanged, the associated squared radius at $\\overline C$ is  \n\\[\nR^{2}(\\overline C)=f\\!\\Bigl(\\tfrac{a+b+c}{3},\\tfrac{a+b+c}{3}\\Bigr)+d^{2}.\n\\]\nBy strict convexity of $f$ and Jensen’s inequality,\n\\[\nf\\!\\Bigl(\\tfrac{a+b+c}{3},\\tfrac{a+b+c}{3}\\Bigr)\n      \\le \\tfrac13\\bigl[f(b,c)+f(c,a)+f(a,b)\\bigr],\n\\]\nwith equality {\\em iff} $(b,c)=(c,a)=(a,b)$, that is, iff $a=b=c$.  \nSince the right–hand side equals $r^{2}-d^{2}$, we obtain  \n\\[\nR^{2}(\\overline C)\\le r^{2},\n\\]\nwith strict inequality unless $a=b=c$.  Hence any radius–minimising centre must satisfy\n\\[\na=b=c.\n\\tag{4}\n\\]\n\n\\medskip\n{\\bf Step 4.  Feasible two–plane of centres.}  \nInserting (4) into (2)–(3) annihilates both constraints, so every point of the two–plane\n\\[\nC=(a,a,a,d),\\qquad a,d\\in\\mathbb R,\n\\tag{5}\n\\]\nis feasible.  On this plane the common squared radius is\n\\[\nr^{2}(a,d)= (a-1)^{2}+a^{2}+d^{2}\n          = 2a^{2}-2a+1+d^{2}.\n\\tag{6}\n\\]\n\n\\medskip\n{\\bf Step 5.  Unconstrained minimisation on the plane.}  \nThe function in (6) is a positive–definite quadratic form in the variables $(a,d)$, hence has a unique global minimiser obtained by solving\n\\[\n\\nabla r^{2}(a,d)=\\bigl(4a-2,\\;2d\\bigr)=\\mathbf 0,\n\\]\nwhich yields\n\\[\na=\\tfrac12,\\qquad d=0.\n\\tag{7}\n\\]\nSubstituting (7) back into (6) gives the minimal squared radius\n\\[\nr_{\\min}^{2}=2\\!\\Bigl(\\tfrac12\\Bigr)^{2}-2\\!\\Bigl(\\tfrac12\\Bigr)+1=\\tfrac12.\n\\tag{8}\n\\]\n\n\\medskip\n{\\bf Step 6.  Verification of uniqueness of the minimiser.}  \nLet $C^{\\ast}=(\\tfrac12,\\tfrac12,\\tfrac12,0)$.  Suppose another feasible centre $\\widetilde C$ attains the same minimal radius.  \nAveraging $\\widetilde C$ over its $G$–orbit cannot increase the radius by the argument of Step 3, so the barycentre is also optimal.  \nBy strict convexity of $f$, equality forces all orbit points to coincide, hence $\\widetilde C=C^{\\ast}$.  Therefore the minimiser is unique.\n\n\\medskip\n{\\bf Step 7.  Equation of the required sphere.}  \nThe unique sphere of smallest radius tangent to $\\ell_{1},\\ell_{2},\\ell_{3}$ is\n\\[\n\\boxed{\\;\n\\bigl(x-\\tfrac12\\bigr)^{2}\n+\\bigl(y-\\tfrac12\\bigr)^{2}\n+\\bigl(z-\\tfrac12\\bigr)^{2}\n+w^{2}\n=\\tfrac12\n\\;},\n\\qquad\nr=\\dfrac{\\sqrt2}{2}.\n\\]\n\n\\medskip\n{\\bf Step 8.  Explicit tangency check (optional).}  \nFor $\\ell_{1}$,\n\\[\n\\lVert C^{\\ast}-(t,1,0,0)\\rVert^{2}\n=(t-\\tfrac12)^{2}+\\bigl(1-\\tfrac12\\bigr)^{2}+\\bigl(0-\\tfrac12\\bigr)^{2}\n=(t-\\tfrac12)^{2}+\\tfrac12 ,\n\\]\nwhose minimum $\\tfrac12$ equals $r^{2}$, confirming tangency.  Cyclic symmetry gives the same result for $\\ell_{2}$ and $\\ell_{3}$.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.517233",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem moves from ℝ³ to ℝ⁴; familiar 3-space tools such as the vector cross-product no longer exist, forcing contestants to use orthogonal–projection formulae and careful linear–algebra manipulation.  \n2. More variables and constraints: three separate tangency conditions (one for each skew line) must hold **simultaneously**, producing a coupled nonlinear system in four unknown coordinates plus the radius.  \n3. Absence of geometric shortcuts: in ℝ³ a shortest common perpendicular gives an immediate construction; in ℝ⁴ no such single segment exists, so an optimisation argument (or Lagrange multipliers) is required.  \n4. Case disjunction and optimisation: after algebraic elimination the solver must recognise and treat two algebraic branches separately and then embed a secondary minimisation to identify the globally minimal radius.  \n5. Symbolic rather than numeric workload: all computations must be carried out exactly, with several quadratic identities and substitutions, precluding mere numerical guessing or pattern matching.\n\nThese added layers make the enhanced variant substantially more intricate and conceptually demanding than either the original examination question or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "In Euclidean four–space $\\mathbb R^{4}$ consider the three pairwise skew affine lines  \n\\[\n\\ell_{1}:\\;(x,y,z,w)=\\bigl(t,\\;1,\\;0,\\;0\\bigr),\\qquad t\\in\\mathbb R,\n\\]\n\\[\n\\ell_{2}:\\;(x,y,z,w)=\\bigl(0,\\;s,\\;1,\\;0\\bigr),\\qquad s\\in\\mathbb R,\n\\]\n\\[\n\\ell_{3}:\\;(x,y,z,w)=\\bigl(1,\\;0,\\;u,\\;0\\bigr),\\qquad u\\in\\mathbb R .\n\\]\n\nAmong all $3$–spheres in $\\mathbb R^{4}$ that are tangent to {\\em each} of the three lines, determine the unique sphere of smallest possible radius.  \nGive its equation in the standard form\n\\[\nS:\\;(x-a)^{2}+(y-b)^{2}+(z-c)^{2}+(w-d)^{2}=r^{2},\n\\]\nand state the centre $(a,b,c,d)$ as well as the radius $r$ in exact radical form.",
      "solution": "{\\bf Step 1.  Squared distances from a general point to the three lines.}  \nFor a point $C=(a,b,c,d)\\in\\mathbb R^{4}$ orthogonal projection onto the lines gives  \n\\[\n\\begin{aligned}\nd_{1}^{2}&=\\operatorname{dist}^{2}(C,\\ell_{1})\n        =(b-1)^{2}+c^{2}+d^{2},\\\\[4pt]\nd_{2}^{2}&=\\operatorname{dist}^{2}(C,\\ell_{2})\n        =a^{2}+(c-1)^{2}+d^{2},\\\\[4pt]\nd_{3}^{2}&=\\operatorname{dist}^{2}(C,\\ell_{3})\n        =(a-1)^{2}+b^{2}+d^{2}.\n\\end{aligned}\n\\]\nIf a $3$–sphere centred at $C$ is tangent simultaneously to $\\ell_{1},\\ell_{2},\\ell_{3}$ we must have\n\\[\nd_{1}^{2}=d_{2}^{2}=d_{3}^{2}=r^{2}.\n\\tag{1}\n\\]\n\n\\medskip\n{\\bf Step 2.  Algebraic form of the tangency constraints.}  \nEquality of the first two expressions in (1) yields\n\\[\n(b-1)^{2}+c^{2}=a^{2}+(c-1)^{2}\n\\;\\Longrightarrow\\;\ng_{1}(a,b,c):=(b-1)^{2}-a^{2}+2c-1=0.\n\\tag{2}\n\\]\nEquality of the first and the third gives\n\\[\n(b-1)^{2}+c^{2}=(a-1)^{2}+b^{2}\n\\;\\Longrightarrow\\;\ng_{2}(a,b,c):=(b-1)^{2}-a^{2}-b^{2}+c^{2}+2a-1=0.\n\\tag{3}\n\\]\nThe coordinate $d$ is unconstrained by (2)–(3).\n\n\\medskip\n{\\bf Step 3.  A symmetry–averaging argument implies $a=b=c$.}  \n\nLet $G=\\langle\\sigma\\rangle\\cong\\mathbb Z_{3}$ act on $\\mathbb R^{4}$ by cyclic permutation  \n\\[\n\\sigma:(x,y,z,w)\\longmapsto(y,z,x,w).\n\\]\nThe unordered set $\\{\\ell_{1},\\ell_{2},\\ell_{3}\\}$ is $G$–invariant.  Hence if a centre $C=(a,b,c,d)$ is feasible, then so are the two points $\\sigma C=(b,c,a,d)$ and $\\sigma^{2}C=(c,a,b,d)$, and all three share the {\\em same} radius $r$ by (1).  \n\nSet \n\\[\nf(b,c):=(b-1)^{2}+c^{2}\\qquad\\bigl(\\text{so }d_{1}^{2}=f(b,c)+d^{2}\\bigr).\n\\]\nThe function $f$ is a strictly convex quadratic form in the pair $(b,c)$; indeed its Hessian is $\\operatorname{diag}(2,2)\\succ 0$.  \nNow average the three orbit points:\n\\[\n\\overline C=\\frac13\\bigl(C+\\sigma C+\\sigma^{2}C\\bigr)\n           =\\Bigl(\\tfrac{a+b+c}{3},\\tfrac{a+b+c}{3},\\tfrac{a+b+c}{3},d\\Bigr).\n\\]\nBecause $d$ is unchanged, the associated squared radius at $\\overline C$ is  \n\\[\nR^{2}(\\overline C)=f\\!\\Bigl(\\tfrac{a+b+c}{3},\\tfrac{a+b+c}{3}\\Bigr)+d^{2}.\n\\]\nBy strict convexity of $f$ and Jensen’s inequality,\n\\[\nf\\!\\Bigl(\\tfrac{a+b+c}{3},\\tfrac{a+b+c}{3}\\Bigr)\n      \\le \\tfrac13\\bigl[f(b,c)+f(c,a)+f(a,b)\\bigr],\n\\]\nwith equality {\\em iff} $(b,c)=(c,a)=(a,b)$, that is, iff $a=b=c$.  \nSince the right–hand side equals $r^{2}-d^{2}$, we obtain  \n\\[\nR^{2}(\\overline C)\\le r^{2},\n\\]\nwith strict inequality unless $a=b=c$.  Hence any radius–minimising centre must satisfy\n\\[\na=b=c.\n\\tag{4}\n\\]\n\n\\medskip\n{\\bf Step 4.  Feasible two–plane of centres.}  \nInserting (4) into (2)–(3) annihilates both constraints, so every point of the two–plane\n\\[\nC=(a,a,a,d),\\qquad a,d\\in\\mathbb R,\n\\tag{5}\n\\]\nis feasible.  On this plane the common squared radius is\n\\[\nr^{2}(a,d)= (a-1)^{2}+a^{2}+d^{2}\n          = 2a^{2}-2a+1+d^{2}.\n\\tag{6}\n\\]\n\n\\medskip\n{\\bf Step 5.  Unconstrained minimisation on the plane.}  \nThe function in (6) is a positive–definite quadratic form in the variables $(a,d)$, hence has a unique global minimiser obtained by solving\n\\[\n\\nabla r^{2}(a,d)=\\bigl(4a-2,\\;2d\\bigr)=\\mathbf 0,\n\\]\nwhich yields\n\\[\na=\\tfrac12,\\qquad d=0.\n\\tag{7}\n\\]\nSubstituting (7) back into (6) gives the minimal squared radius\n\\[\nr_{\\min}^{2}=2\\!\\Bigl(\\tfrac12\\Bigr)^{2}-2\\!\\Bigl(\\tfrac12\\Bigr)+1=\\tfrac12.\n\\tag{8}\n\\]\n\n\\medskip\n{\\bf Step 6.  Verification of uniqueness of the minimiser.}  \nLet $C^{\\ast}=(\\tfrac12,\\tfrac12,\\tfrac12,0)$.  Suppose another feasible centre $\\widetilde C$ attains the same minimal radius.  \nAveraging $\\widetilde C$ over its $G$–orbit cannot increase the radius by the argument of Step 3, so the barycentre is also optimal.  \nBy strict convexity of $f$, equality forces all orbit points to coincide, hence $\\widetilde C=C^{\\ast}$.  Therefore the minimiser is unique.\n\n\\medskip\n{\\bf Step 7.  Equation of the required sphere.}  \nThe unique sphere of smallest radius tangent to $\\ell_{1},\\ell_{2},\\ell_{3}$ is\n\\[\n\\boxed{\\;\n\\bigl(x-\\tfrac12\\bigr)^{2}\n+\\bigl(y-\\tfrac12\\bigr)^{2}\n+\\bigl(z-\\tfrac12\\bigr)^{2}\n+w^{2}\n=\\tfrac12\n\\;},\n\\qquad\nr=\\dfrac{\\sqrt2}{2}.\n\\]\n\n\\medskip\n{\\bf Step 8.  Explicit tangency check (optional).}  \nFor $\\ell_{1}$,\n\\[\n\\lVert C^{\\ast}-(t,1,0,0)\\rVert^{2}\n=(t-\\tfrac12)^{2}+\\bigl(1-\\tfrac12\\bigr)^{2}+\\bigl(0-\\tfrac12\\bigr)^{2}\n=(t-\\tfrac12)^{2}+\\tfrac12 ,\n\\]\nwhose minimum $\\tfrac12$ equals $r^{2}$, confirming tangency.  Cyclic symmetry gives the same result for $\\ell_{2}$ and $\\ell_{3}$.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.432979",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem moves from ℝ³ to ℝ⁴; familiar 3-space tools such as the vector cross-product no longer exist, forcing contestants to use orthogonal–projection formulae and careful linear–algebra manipulation.  \n2. More variables and constraints: three separate tangency conditions (one for each skew line) must hold **simultaneously**, producing a coupled nonlinear system in four unknown coordinates plus the radius.  \n3. Absence of geometric shortcuts: in ℝ³ a shortest common perpendicular gives an immediate construction; in ℝ⁴ no such single segment exists, so an optimisation argument (or Lagrange multipliers) is required.  \n4. Case disjunction and optimisation: after algebraic elimination the solver must recognise and treat two algebraic branches separately and then embed a secondary minimisation to identify the globally minimal radius.  \n5. Symbolic rather than numeric workload: all computations must be carried out exactly, with several quadratic identities and substitutions, precluding mere numerical guessing or pattern matching.\n\nThese added layers make the enhanced variant substantially more intricate and conceptually demanding than either the original examination question or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}