{
  "index": "1940-A-4",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "4. The parabola \\( y^{2}=-4 p x \\) rolls without slipping around the parabola \\( y^{2}=4 p x \\). Find the equation of the locus of the vertex of the rolling parabola.",
  "solution": "Solution. If the rolling parabola and the fixed parabola are tangent at the point \\( Q \\), it is obvious from symmetry that the vertex \\( V \\) of the rolling parabola is the reflection of the origin (the vertex of the fixed parabola) in the tangent line at \\( Q \\).\nIn the sketch, we have tacitly assumed \\( p>0 \\). Suppose that \\( Q \\) is the point ( \\( 4 p t^{2}, 4 p t \\) ). [Any point on the fixed parabola will have this form for a unique \\( t \\).] The slope of the tangent at \\( Q \\) is \\( 1 /(2 t) \\) and the equation of the tangent is\n\\[\ny=\\frac{1}{2 t} x+2 p t .\n\\]\n\nThe perpendicular on this line through the origin has the equation\n\\[\ny=-2 t x .\n\\]\n\nThese lines intersect at\n\\[\n\\left(\\frac{-4 p t^{2}}{1+4 t^{2}}, \\frac{8 p t^{3}}{1+4 t^{2}}\\right)\n\\]\nso the vertex \\( V \\) is at\n\\[\n\\left(\\frac{-8 p t^{2}}{1+4 t^{2}}, \\frac{16 p t^{3}}{1+4 t^{2}}\\right) .\n\\]\n\nThis gives us the parametric equations\n\\[\nx=\\frac{-8 p t^{2}}{1+4 t^{2}}, \\quad y=\\frac{16 p t^{3}}{1+4 t^{2}}\n\\]\nfor the path of the point \\( V \\). Since \\( -2 t=y / x \\) (if \\( x \\neq 0) \\), the elimination of \\( t \\) gives\n\\[\nx=\\frac{-2 p\\left(\\frac{y}{x}\\right)^{2}}{1+\\left(\\frac{y}{x}\\right)^{2}}\n\\]\nso that\n\\[\n\\left(x^{2}+y^{2}\\right) x+2 p y^{2}=0\n\\]\nis an equation satisfied by all points of the locus (including \\( (0,0) \\) ). Conversely, any point \\( (x, y) \\) other than the origin satisfying (1) leads to a value of \\( t(=-y / 2 x) \\) and is therefore on the locus. Thus (1) describes the locus precisely.",
  "vars": [
    "x",
    "y",
    "t",
    "Q",
    "V"
  ],
  "params": [
    "p"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "t": "parameter",
        "Q": "contactpoint",
        "V": "rollingvertex",
        "p": "focusdist"
      },
      "question": "4. The parabola \\( ordinate^{2}=-4 focusdist abscissa \\) rolls without slipping around the parabola \\( ordinate^{2}=4 focusdist abscissa \\). Find the equation of the locus of the vertex of the rolling parabola.",
      "solution": "Solution. If the rolling parabola and the fixed parabola are tangent at the point \\( contactpoint \\), it is obvious from symmetry that the vertex \\( rollingvertex \\) of the rolling parabola is the reflection of the origin (the vertex of the fixed parabola) in the tangent line at \\( contactpoint \\).\nIn the sketch, we have tacitly assumed \\( focusdist>0 \\). Suppose that \\( contactpoint \\) is the point ( \\( 4 focusdist parameter^{2}, 4 focusdist parameter \\) ). [Any point on the fixed parabola will have this form for a unique \\( parameter \\).] The slope of the tangent at \\( contactpoint \\) is \\( 1 /(2 parameter) \\) and the equation of the tangent is\n\\[\nordinate=\\frac{1}{2 parameter} abscissa+2 focusdist parameter .\n\\]\n\nThe perpendicular on this line through the origin has the equation\n\\[\nordinate=-2 parameter abscissa .\n\\]\n\nThese lines intersect at\n\\[\n\\left(\\frac{-4 focusdist parameter^{2}}{1+4 parameter^{2}}, \\frac{8 focusdist parameter^{3}}{1+4 parameter^{2}}\\right)\n\\]\nso the vertex \\( rollingvertex \\) is at\n\\[\n\\left(\\frac{-8 focusdist parameter^{2}}{1+4 parameter^{2}}, \\frac{16 focusdist parameter^{3}}{1+4 parameter^{2}}\\right) .\n\\]\n\nThis gives us the parametric equations\n\\[\nabscissa=\\frac{-8 focusdist parameter^{2}}{1+4 parameter^{2}}, \\quad ordinate=\\frac{16 focusdist parameter^{3}}{1+4 parameter^{2}}\n\\]\nfor the path of the point \\( rollingvertex \\). Since \\( -2 parameter=ordinate / abscissa \\) (if \\( abscissa \\neq 0) \\), the elimination of \\( parameter \\) gives\n\\[\nabscissa=\\frac{-2 focusdist\\left(\\frac{ordinate}{abscissa}\\right)^{2}}{1+\\left(\\frac{ordinate}{abscissa}\\right)^{2}}\n\\]\nso that\n\\[\n\\left(abscissa^{2}+ordinate^{2}\\right) abscissa+2 focusdist ordinate^{2}=0\n\\]\nis an equation satisfied by all points of the locus (including \\( (0,0) \\) ). Conversely, any point \\( (abscissa, ordinate) \\) other than the origin satisfying (1) leads to a value of \\( parameter(=-ordinate / 2 abscissa) \\) and is therefore on the locus. Thus (1) describes the locus precisely."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "y": "dragonfly",
        "t": "butterscotch",
        "Q": "hummingbird",
        "V": "chandeliers",
        "p": "raspberries"
      },
      "question": "4. The parabola \\( dragonfly^{2}=-4 raspberries sunflower \\) rolls without slipping around the parabola \\( dragonfly^{2}=4 raspberries sunflower \\). Find the equation of the locus of the vertex of the rolling parabola.",
      "solution": "Solution. If the rolling parabola and the fixed parabola are tangent at the point \\( hummingbird \\), it is obvious from symmetry that the vertex \\( chandeliers \\) of the rolling parabola is the reflection of the origin (the vertex of the fixed parabola) in the tangent line at \\( hummingbird \\).\nIn the sketch, we have tacitly assumed \\( raspberries>0 \\). Suppose that \\( hummingbird \\) is the point ( \\( 4 raspberries butterscotch^{2}, 4 raspberries butterscotch \\) ). [Any point on the fixed parabola will have this form for a unique \\( butterscotch \\).] The slope of the tangent at \\( hummingbird \\) is \\( 1 /(2 butterscotch) \\) and the equation of the tangent is\n\\[\ndragonfly=\\frac{1}{2 butterscotch} sunflower+2 raspberries butterscotch .\n\\]\n\nThe perpendicular on this line through the origin has the equation\n\\[\ndragonfly=-2 butterscotch sunflower .\n\\]\n\nThese lines intersect at\n\\[\n\\left(\\frac{-4 raspberries butterscotch^{2}}{1+4 butterscotch^{2}}, \\frac{8 raspberries butterscotch^{3}}{1+4 butterscotch^{2}}\\right)\n\\]\nso the vertex \\( chandeliers \\) is at\n\\[\n\\left(\\frac{-8 raspberries butterscotch^{2}}{1+4 butterscotch^{2}}, \\frac{16 raspberries butterscotch^{3}}{1+4 butterscotch^{2}}\\right) .\n\\]\n\nThis gives us the parametric equations\n\\[\nsunflower=\\frac{-8 raspberries butterscotch^{2}}{1+4 butterscotch^{2}}, \\quad dragonfly=\\frac{16 raspberries butterscotch^{3}}{1+4 butterscotch^{2}}\n\\]\nfor the path of the point \\( chandeliers \\). Since \\( -2 butterscotch=dragonfly / sunflower \\) (if \\( sunflower \\neq 0) \\), the elimination of \\( butterscotch \\) gives\n\\[\nsunflower=\\frac{-2 raspberries\\left(\\frac{dragonfly}{sunflower}\\right)^{2}}{1+\\left(\\frac{dragonfly}{sunflower}\\right)^{2}}\n\\]\nso that\n\\[\n\\left(sunflower^{2}+dragonfly^{2}\\right) sunflower+2 raspberries dragonfly^{2}=0\n\\]\nis an equation satisfied by all points of the locus (including \\( (0,0) \\) ). Conversely, any point \\( (sunflower, dragonfly) \\) other than the origin satisfying (1) leads to a value of \\( butterscotch(=-dragonfly / 2 sunflower) \\) and is therefore on the locus. Thus (1) describes the locus precisely."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "t": "stagnation",
        "Q": "separationpoint",
        "V": "troughpoint",
        "p": "variability"
      },
      "question": "4. The parabola \\( horizontalaxis^{2}=-4 variability verticalaxis \\) rolls without slipping around the parabola \\( horizontalaxis^{2}=4 variability verticalaxis \\). Find the equation of the locus of the vertex of the rolling parabola.",
      "solution": "Solution. If the rolling parabola and the fixed parabola are tangent at the point \\( separationpoint \\), it is obvious from symmetry that the vertex \\( troughpoint \\) of the rolling parabola is the reflection of the origin (the vertex of the fixed parabola) in the tangent line at \\( separationpoint \\).\nIn the sketch, we have tacitly assumed \\( variability>0 \\). Suppose that \\( separationpoint \\) is the point ( \\( 4 variability stagnation^{2}, 4 variability stagnation \\) ). [Any point on the fixed parabola will have this form for a unique \\( stagnation \\).] The slope of the tangent at \\( separationpoint \\) is \\( 1 /(2 stagnation) \\) and the equation of the tangent is\n\\[\nhorizontalaxis=\\frac{1}{2 stagnation} verticalaxis+2 variability stagnation .\n\\]\n\nThe perpendicular on this line through the origin has the equation\n\\[\nhorizontalaxis=-2 stagnation verticalaxis .\n\\]\n\nThese lines intersect at\n\\[\n\\left(\\frac{-4 variability stagnation^{2}}{1+4 stagnation^{2}}, \\frac{8 variability stagnation^{3}}{1+4 stagnation^{2}}\\right)\n\\]\nso the vertex \\( troughpoint \\) is at\n\\[\n\\left(\\frac{-8 variability stagnation^{2}}{1+4 stagnation^{2}}, \\frac{16 variability stagnation^{3}}{1+4 stagnation^{2}}\\right) .\n\\]\n\nThis gives us the parametric equations\n\\[\nverticalaxis=\\frac{-8 variability stagnation^{2}}{1+4 stagnation^{2}}, \\quad horizontalaxis=\\frac{16 variability stagnation^{3}}{1+4 stagnation^{2}}\n\\]\nfor the path of the point \\( troughpoint \\). Since \\( -2 stagnation=horizontalaxis / verticalaxis \\) (if \\( verticalaxis \\neq 0) \\), the elimination of \\( stagnation \\) gives\n\\[\nverticalaxis=\\frac{-2 variability\\left(\\frac{horizontalaxis}{verticalaxis}\\right)^{2}}{1+\\left(\\frac{horizontalaxis}{verticalaxis}\\right)^{2}}\n\\]\nso that\n\\[\n\\left(verticalaxis^{2}+horizontalaxis^{2}\\right) verticalaxis+2 variability horizontalaxis^{2}=0\n\\]\nis an equation satisfied by all points of the locus (including \\( (0,0) \\) ). Conversely, any point \\( (verticalaxis, horizontalaxis) \\) other than the origin satisfying (1) leads to a value of \\( stagnation(=-horizontalaxis / 2 verticalaxis) \\) and is therefore on the locus. Thus (1) describes the locus precisely."
    },
    "garbled_string": {
      "map": {
        "x": "navqgero",
        "y": "qlzmtvha",
        "t": "jxfrilud",
        "Q": "zpsowknh",
        "V": "guxderib",
        "p": "lykovema"
      },
      "question": "4. The parabola \\( qlzmtvha^{2}=-4 lykovema navqgero \\) rolls without slipping around the parabola \\( qlzmtvha^{2}=4 lykovema navqgero \\). Find the equation of the locus of the vertex of the rolling parabola.",
      "solution": "Solution. If the rolling parabola and the fixed parabola are tangent at the point \\( zpsowknh \\), it is obvious from symmetry that the vertex \\( guxderib \\) of the rolling parabola is the reflection of the origin (the vertex of the fixed parabola) in the tangent line at \\( zpsowknh \\).\nIn the sketch, we have tacitly assumed \\( lykovema>0 \\). Suppose that \\( zpsowknh \\) is the point ( \\( 4 lykovema jxfrilud^{2}, 4 lykovema jxfrilud \\) ). [Any point on the fixed parabola will have this form for a unique \\( jxfrilud \\).] The slope of the tangent at \\( zpsowknh \\) is \\( 1 /(2 jxfrilud) \\) and the equation of the tangent is\n\\[\nqlzmtvha=\\frac{1}{2 jxfrilud} navqgero+2 lykovema jxfrilud .\n\\]\n\nThe perpendicular on this line through the origin has the equation\n\\[\nqlzmtvha=-2 jxfrilud navqgero .\n\\]\n\nThese lines intersect at\n\\[\n\\left(\\frac{-4 lykovema jxfrilud^{2}}{1+4 jxfrilud^{2}}, \\frac{8 lykovema jxfrilud^{3}}{1+4 jxfrilud^{2}}\\right)\n\\]\nso the vertex \\( guxderib \\) is at\n\\[\n\\left(\\frac{-8 lykovema jxfrilud^{2}}{1+4 jxfrilud^{2}}, \\frac{16 lykovema jxfrilud^{3}}{1+4 jxfrilud^{2}}\\right) .\n\\]\n\nThis gives us the parametric equations\n\\[\nnavqgero=\\frac{-8 lykovema jxfrilud^{2}}{1+4 jxfrilud^{2}}, \\quad qlzmtvha=\\frac{16 lykovema jxfrilud^{3}}{1+4 jxfrilud^{2}}\n\\]\nfor the path of the point \\( guxderib \\). Since \\( -2 jxfrilud=qlzmtvha / navqgero \\) (if \\( navqgero \\neq 0) \\), the elimination of \\( jxfrilud \\) gives\n\\[\nnavqgero=\\frac{-2 lykovema\\left(\\frac{qlzmtvha}{navqgero}\\right)^{2}}{1+\\left(\\frac{qlzmtvha}{navqgero}\\right)^{2}}\n\\]\nso that\n\\[\n\\left(navqgero^{2}+qlzmtvha^{2}\\right) navqgero+2 lykovema qlzmtvha^{2}=0\n\\]\nis an equation satisfied by all points of the locus (including \\( (0,0) \\) ). Conversely, any point \\( (navqgero, qlzmtvha) \\) other than the origin satisfying (1) leads to a value of \\( jxfrilud(=-qlzmtvha / 2 navqgero) \\) and is therefore on the locus. Thus (1) describes the locus precisely."
    },
    "kernel_variant": {
      "question": "Let $p>0$.\n\n(1) In the fixed Euclidean space $\\mathbb R^{3}$ consider the upward–opening circular paraboloid  \n\\[\n\\Sigma_{1}\\colon\\;z=\\dfrac{x^{2}+y^{2}}{4p}.\n\\]\n\n(2) A congruent downward–opening paraboloid, expressed in its own body–fixed coordinates by  \n\\[\n\\widehat\\Sigma_{2}\\colon\\;\\hat z=-\\dfrac{\\hat x^{2}+\\hat y^{2}}{4p},\n\\]\nis put in exterior contact with $\\Sigma_{1}$ and then allowed to move on $\\Sigma_{1}$  \n\n\\qquad$\\bullet$ without slipping (the linear velocities of the two bodies coincide at the\ninstantaneous point of contact),  \n\n\\qquad$\\bullet$ without twisting (the relative spin about the common normal vanishes).\n\nInitial configuration.  \nChoose the meridian plane $y=0$.  At $t=0$ the two paraboloids touch at the point  \n\\[\nQ_{0}=(2p,0,p)\\in\\Sigma_{1}.\n\\]\nThe tangent plane to $\\Sigma_{1}$ at $Q_{0}$ has the equation $z=x-p$.  Reflect the origin $O$ (the vertex of $\\Sigma_{1}$) in this plane and denote the image by  \n\\[\nV_{0}=(p,0,-p).\n\\]\nPlace the vertex $V_{0}$ of the moving paraboloid (now called $\\Sigma_{2}$) at this point and orient $\\Sigma_{2}$ so that  \n\n\\qquad$\\bullet$ it is tangent to $\\Sigma_{1}$ at $Q_{0}$, and  \n\n\\qquad$\\bullet$ its symmetry axis $A_{2}(0)$ lies in the plane $y=0$,  \n\nso that $\\Sigma_{2}$ is entirely outside $\\Sigma_{1}$.  \nThe motion is then continued, still satisfying the no–slip/no–twist constraints, until $\\Sigma_{2}$ has returned to its initial attitude after one complete revolution about the $z$–axis.\n\n(a)  For every instant $t$ let $\\Pi(t)$ be the unique meridian plane containing the $z$–axis and the current contact point $Q(t)$.  \n\n(i) Prove that the symmetry axis $A_{2}(t)$ of $\\Sigma_{2}$ is contained in $\\Pi(t)$.  \n\n(ii) Let $\\boldsymbol\\omega(t)$ be the angular–velocity vector of $\\Sigma_{2}$ relative to $\\Sigma_{1}$.  Show that  \n\n\\[\n\\boldsymbol\\omega(t)\\in\\Pi(t)\\quad\\Longleftrightarrow\\quad\n\\bigl(\\forall P\\in\\Pi(t)\\bigr)\\,v_{P}(t)\\perp\\Pi(t),\n\\]\n\nand, making full use of the rolling constraints and the symmetries of the configuration, deduce that  \n\\[\n\\boxed{\\boldsymbol\\omega(t)\\in\\Pi(t)\\quad\\text{and}\\quad\n       \\boldsymbol\\omega(t)\\parallel\\mathbf e_{r}(t)},\n\\]\nthat is, $\\boldsymbol\\omega(t)$ is parallel to the radial principal direction.\n\n(b)  Restrict the study to the fixed meridian plane $\\Pi(0)=\\{y=0\\}$.  While the contact point $Q(t)$ remains in this plane, determine the planar locus $\\ell$ traced by the vertex $V(t)$ of $\\Sigma_{2}$ under the rolling constraints.  Show further that, when $\\Pi(t)$ is allowed to rotate with the motion, the full three–dimensional locus $\\mathcal L$ traced by $V$ during one complete revolution is the cubic surface  \n\n\\[\nF(x,y,z):=(x^{2}+y^{2}+z^{2})\\,z+2p\\,(x^{2}+y^{2})=0.\n\\]\n\nState explicitly which branch of $\\mathcal L$ is actually generated by the motion and justify every geometric step that leads from the planar curve $\\ell$ to the surface $F=0$.\n\n(c)  Determine the algebraic degree of $\\mathcal L$ and verify it directly from the equation $F=0$.",
      "solution": "Throughout boldface symbols denote vectors in $\\mathbb R^{3}$, ordinary letters their\ncoordinates in the fixed frame $(O;x,y,z)$.  \nAt the contact point $Q(t)$ put  \n\\[\n\\mathbf n(t)\\;\\text{(unit normal pointing outward from }\\Sigma_{1}),\\quad\n\\mathbf e_{r}(t)\\;\\text{(radial unit vector in }T_{Q(t)}\\Sigma_{1}),\\quad\n\\mathbf e_{\\theta}(t)=\\mathbf n(t)\\times\\mathbf e_{r}(t).\n\\]\nThe ordered triple $\\bigl(\\mathbf e_{r},\\mathbf e_{\\theta},\\mathbf n\\bigr)$ is direct and orthonormal.\n\n-----------------------------------------------------------------\n1.  Part (a)(i): the axis $A_{2}(t)$ lies in $\\Pi(t)$  \n-----------------------------------------------------------------\n\nExactly as in any surface–of–revolution argument, $\\mathbf e_{\\theta}(t)$ is the \\emph{unique} azimuthal principal direction at $Q(t)$.  \nBecause the no–twist constraint forces the principal frames of the two\ncongruent surfaces to coincide, the meridian direction of $\\Sigma_{2}(t)$ must\nbe the line orthogonal to $\\mathbf e_{\\theta}(t)$, namely $\\Pi(t)$.  Since the\nsymmetry axis of a surface of revolution belongs to each of its meridian\nplanes, we obtain\n\\[\n\\boxed{A_{2}(t)\\subset\\Pi(t)\\quad\\forall\\,t}.\n\\]\n\n-----------------------------------------------------------------\n2.  Part (a)(ii): direction of the angular–velocity vector  \n-----------------------------------------------------------------\n\n-----------------------------------------------------------------\n2.1  A correct proof of the key equivalence  \n-----------------------------------------------------------------  \n\nLet $\\Pi(t)$ be a plane through the origin\\footnote{Only the direction of the plane matters for the present argument; translating it by the contact point would not change anything.} with unit normal $\\mathbf m(t)=\\mathbf e_{\\theta}(t)$.  \nDecompose an arbitrary vector $\\boldsymbol\\omega$ into its normal and tangential components:\n\\[\n\\boldsymbol\\omega=\\omega_{\\perp}\\,\\mathbf m+\\boldsymbol\\omega_{\\parallel},\n\\qquad\\boldsymbol\\omega_{\\parallel}\\in\\Pi(t).\n\\]\nFor every $\\mathbf u\\in\\Pi(t)$ the linear velocity of the material point whose (instantaneous) position vector is $\\mathbf u$ reads\n\\[\n\\mathbf v(\\mathbf u)=\\boldsymbol\\omega\\times\\mathbf u.\n\\]\nTake any $\\mathbf w\\in\\Pi(t)$.  Using the scalar triple product,\n\\[\n\\mathbf v(\\mathbf u)\\cdot\\mathbf w\n       =(\\boldsymbol\\omega\\times\\mathbf u)\\cdot\\mathbf w\n       =\\boldsymbol\\omega\\cdot(\\mathbf u\\times\\mathbf w).\n\\]\nBecause $\\mathbf u,\\mathbf w\\in\\Pi(t)$, their cross–product $\\mathbf u\\times\\mathbf w$ is parallel to $\\mathbf m$.  Hence\n\\[\n\\mathbf v(\\mathbf u)\\cdot\\mathbf w\n       =\\omega_{\\perp}\\,\\mathbf m\\cdot(\\mathbf u\\times\\mathbf w).\n\\]\nTherefore  \n\n$\\displaystyle\\bigl(\\forall\\mathbf u,\\mathbf w\\in\\Pi(t)\\bigr)\\;\n         \\mathbf v(\\mathbf u)\\cdot\\mathbf w=0\n\\quad\\Longleftrightarrow\\quad\n         \\omega_{\\perp}=0\n\\quad\\Longleftrightarrow\\quad\n         \\boldsymbol\\omega\\in\\Pi(t).$\n\nIn words,\n\\[\n\\boxed{\\;\n\\boldsymbol\\omega\\in\\Pi(t)\n\\;\\Longleftrightarrow\\;\n\\bigl(\\forall P\\in\\Pi(t)\\bigr)\\,v_{P}\\perp\\Pi(t)\n\\;}\n\\tag{2.1}\n\\]\nwhich is the desired equivalence.\n\n-----------------------------------------------------------------\n2.2  Elimination of the normal component  \n-----------------------------------------------------------------\n\nThe no–twist constraint literally says that the relative spin about the common\nnormal is zero, so that  \n\\[\n\\omega_{n}:=\\boldsymbol\\omega\\cdot\\mathbf n(t)=0. \\tag{2.2}\n\\]\nTherefore $\\boldsymbol\\omega$ already belongs to $T_{Q(t)}\\Sigma_{1}$.  It\nremains to rule out a component along $\\mathbf e_{\\theta}(t)$.\n\n-----------------------------------------------------------------\n2.3  A symmetry argument that forces $\\omega_{\\theta}=0$  \n-----------------------------------------------------------------\n\nThe entire configuration at time $t$ is\n\\emph{mirror–symmetric} with respect to $\\Pi(t)$:\n\n(i)  The fixed paraboloid $\\Sigma_{1}$ is invariant under the reflection  \n\\[\n\\mathscr R_{\\Pi(t)}\\colon(x,y,z)\\longmapsto(x,-y,z).\n\\]\n\n(ii)  $\\Sigma_{2}(t)$ shares that symmetry, because its\nvertex $V(t)$ and its axis $A_{2}(t)$ both lie in $\\Pi(t)$ (proved in\npart (a)(i)).\n\n(iii)  The non–holonomic constraints (no–slip, no–twist) are geometric\nconditions preserved by every ambient isometry, in particular by\n$\\mathscr R_{\\Pi(t)}$.\n\nHence, if $\\bigl(\\Sigma_{2}(t),\\boldsymbol\\omega(t)\\bigr)$ is an admissible\nstate, so is  \n$\\bigl(\\mathscr R_{\\Pi(t)}\\Sigma_{2}(t),\\,\n       \\mathscr R_{\\Pi(t)\\,*}\\boldsymbol\\omega(t)\\bigr)$.\nBut by (ii) the first components coincide, therefore\n\\[\n\\mathscr R_{\\Pi(t)\\,*}\\boldsymbol\\omega(t)=\\boldsymbol\\omega(t).\n\\tag{2.3}\n\\]\n\nNow $\\mathscr R_{\\Pi(t)}$ fixes $\\mathbf e_{r}(t)$ and $\\mathbf n(t)$ but\n\\emph{reverses the sign} of $\\mathbf e_{\\theta}(t)$.  Writing\n\\[\n\\boldsymbol\\omega=\\omega_{r}\\,\\mathbf e_{r}\n                 +\\omega_{\\theta}\\,\\mathbf e_{\\theta},\n\\tag{2.4}\n\\]\nrelation \\eqref{2.3} gives\n\\[\n\\mathscr R_{\\Pi(t)\\,*}\\boldsymbol\\omega\n     =\\omega_{r}\\,\\mathbf e_{r}-\\omega_{\\theta}\\,\\mathbf e_{\\theta}\n     =\\boldsymbol\\omega\n     \\;\\Longrightarrow\\;\n     \\boxed{\\omega_{\\theta}=0}. \\tag{2.5}\n\\]\n\n-----------------------------------------------------------------\n2.4  Conclusion for the angular velocity  \n-----------------------------------------------------------------\n\nCombining \\eqref{2.2} and \\eqref{2.5},\n\\[\n\\boxed{\\boldsymbol\\omega(t)=\\omega_{r}(t)\\,\\mathbf e_{r}(t)},\\qquad\n\\boldsymbol\\omega(t)\\in\\Pi(t)\\;\\text{and}\\;\n\\boldsymbol\\omega(t)\\perp\\mathbf e_{\\theta}(t),\n\\]\nwhich simultaneously proves the boxed statement required in part (a)(ii).\n\n-----------------------------------------------------------------\n3.  Geometry inside the fixed meridian plane $\\Pi(0)=\\{y=0\\}$  \n-----------------------------------------------------------------\n\nSuppress the $y$–coordinate and rename $x=r,\\;z=z$.  \nIntersecting the paraboloids with the plane gives the congruent parabolas  \n\\[\nC_{1}\\colon\\;z=\\dfrac{r^{2}}{4p},\\qquad\nC_{2}\\colon\\;z=-\\dfrac{r^{2}}{4p}. \\tag{3.1}\n\\]\n\nParameterise the fixed parabola $C_{1}$ by the focal parameter $t>0$:\n\\[\nQ(t)=(r(t),z(t))=(2pt,\\;pt^{2}). \\tag{3.2}\n\\]\nThe tangent line $\\tau(t)$ at $Q(t)$ has slope $dz/dr=r/(2p)=t$; hence  \n\\[\n\\tau(t)\\colon\\;z-pt^{2}=t\\,(r-2pt). \\tag{3.3}\n\\]\n\nBecause the parabolas are congruent, $\\tau(t)$ is the axis of\nsymmetry for the ordered pair $(C_{1},C_{2})$.  Consequently the vertex\n$v(t)$ of $C_{2}$ is the reflection of the origin $O=(0,0)$ in $\\tau(t)$.\n\nFoot of the perpendicular.  \nLet $H(t)$ be the foot of the perpendicular from $O$ onto $\\tau(t)$.  Solving\n\\eqref{3.3} together with $z=-(1/t)\\,r$ yields  \n\\[\nH(t)=\\Bigl(\\dfrac{pt^{3}}{1+t^{2}},\\;-\\dfrac{pt^{2}}{1+t^{2}}\\Bigr). \\tag{3.4}\n\\]\n\nReflection formula.  \nTherefore  \n\\[\nv(t)=2H(t)-O\n     =\\Bigl(\\dfrac{2pt^{3}}{1+t^{2}},\\;-\\dfrac{2pt^{2}}{1+t^{2}}\\Bigr). \\tag{3.5}\n\\]\n\nAt $t=1$ one finds $v(1)=(p,-p)$, agreeing with the prescribed initial vertex\n$V_{0}$.\n\n-----------------------------------------------------------------\n4.  Implicit equation of the planar locus $\\ell$  \n-----------------------------------------------------------------\n\nEliminate $t$ from \\eqref{3.5}.  From the first component one gets\n$t=r_{v}/(-z_{v})$ (because $z_{v}<0$).  Substituting into the second\ncomponent gives  \n\\[\nz_{v}\\,(r_{v}^{2}+z_{v}^{2})+2p\\,r_{v}^{2}=0. \\tag{4.1}\n\\]\nThus  \n\\[\n\\boxed{\\ \\ell=\\bigl\\{(r,z)\\in\\mathbb R^{2}\\colon z(r^{2}+z^{2})+2p\\,r^{2}=0\\bigr\\}\\ }. \\tag{4.2}\n\\]\n\nPhysical branch.  \nDuring the motion $t>0\\;\\Rightarrow\\;r_{v}>0,\\;z_{v}<0$; hence only the branch\n$z<0$ of the cubic \\eqref{4.2} is realised.\n\n-----------------------------------------------------------------\n5.  Generation of the spatial locus $\\mathcal L$  \n-----------------------------------------------------------------\n\nBy part (a)(i) the contact point $Q(t)$ and the vertex $V(t)$ lie in the\n\\emph{same} meridian plane $\\Pi(t)$.  Consequently they share the same\nazimuthal angle $\\varphi(t)$.  Rotating the planar curve $\\ell$ about the\n$z$–axis by that angle sweeps the full spatial locus:\n\\[\n\\mathcal L=\\bigl\\{(r\\cos\\varphi,\\;r\\sin\\varphi,\\;z)\\colon\\;(r,z)\\in\\ell,\\;\n             \\varphi\\in[0,2\\pi)\\bigr\\}.\n\\]\nReplacing $r$ by $\\sqrt{x^{2}+y^{2}}$ in \\eqref{4.1} gives the Cartesian\nimplicit polynomial  \n\\[\nF(x,y,z):=(x^{2}+y^{2}+z^{2})\\,z+2p\\,(x^{2}+y^{2})=0,\n\\]\nwhich is exactly the surface announced in the problem.\n\nBranch realised.  \nBecause the motion keeps $z<0$, the physically attainable part is  \n\\[\n\\mathcal L_{\\text{phys}}=\\mathcal L\\cap\\{\\,z<0\\,\\}.\n\\]\n(The algebraic point $(0,0,0)$ satisfies $F=0$ but is never reached.)\n\n-----------------------------------------------------------------\n6.  Part (c): algebraic degree of $\\mathcal L$  \n-----------------------------------------------------------------\n\nThe polynomial $F$ contains the monomials $z^{3}$ and $z(x^{2}+y^{2})$, both\nof total degree $3$, and no term of higher degree.  Hence $\\mathcal L$ is a\ncubic surface of revolution.  Its intersection with the plane $y=0$ reproduces\n\\eqref{4.2}, confirming the degree once more.\n\n-----------------------------------------------------------------\n7.  Closure of the motion  \n-----------------------------------------------------------------\n\nWhen the azimuth $\\varphi(t)$ increases by $2\\pi$, the moving paraboloid\n$\\Sigma_{2}$ (hence its vertex) returns to its initial attitude; the branch\n$z<0$ of the cubic surface is therefore described exactly once during a full\nrevolution.\n\n-----------------------------------------------------------------\nAnswer.  \nThe vertex of the rolling paraboloid traces precisely the branch $z<0$ of the\ncubic surface  \n\\[\n\\boxed{(x^{2}+y^{2}+z^{2})\\,z+2p\\,(x^{2}+y^{2})=0}.\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.369351",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension – The original problem is planar; the enhanced variant takes place in 3-space, demanding the passage from a curve to a surface and an understanding of surfaces of revolution.\n\n2. Additional structures – Rolling contact between two 2-dimensional surfaces introduces differential-geometric ideas (common normal, no-slip kinematics) that are absent in the 2-D case.\n\n3. Reduction & symmetry arguments – A successful solution requires reducing the 3-D motion to a 2-D meridian problem via rotational symmetry, then lifting the answer back to 3-D.  Handling this reduction rigorously involves group actions and invariance considerations.\n\n4. More intricate algebra – Eliminating the parameter now produces a quartic surface equation instead of a cubic curve, and the algebraic manipulation is appreciably heavier.\n\n5. Conceptual depth – Understanding why the vertex is the reflection of O across the tangent in three dimensions (it is really the reflection in the tangent plane along the normal) calls for familiarity with reflections in planes and with tangent planes to surfaces.\n\nOverall, the enhanced kernel variant requires geometric insight in higher dimensions, facility with differential geometry of surfaces, exploitation of symmetry groups, and more elaborate algebraic elimination, making it substantially harder than both the original and the given kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $p>0$.\n\n(1) In the fixed Euclidean space $\\mathbb R^{3}$ consider the upward–opening circular paraboloid  \n\\[\n\\Sigma_{1}\\colon\\;z=\\dfrac{x^{2}+y^{2}}{4p}.\n\\]\n\n(2) A congruent downward–opening paraboloid, expressed in its own body–fixed coordinates by  \n\\[\n\\widehat\\Sigma_{2}\\colon\\;\\hat z=-\\dfrac{\\hat x^{2}+\\hat y^{2}}{4p},\n\\]\nis put in exterior contact with $\\Sigma_{1}$ and then allowed to move on $\\Sigma_{1}$  \n\n\\qquad$\\bullet$ without slipping (the linear velocities of the two bodies coincide at the\ninstantaneous point of contact),  \n\n\\qquad$\\bullet$ without twisting (the relative spin about the common normal vanishes).\n\nInitial configuration.  \nChoose the meridian plane $y=0$.  At $t=0$ the two paraboloids touch at the point  \n\\[\nQ_{0}=(2p,0,p)\\in\\Sigma_{1}.\n\\]\nThe tangent plane to $\\Sigma_{1}$ at $Q_{0}$ has the equation $z=x-p$.  Reflect the origin $O$ (the vertex of $\\Sigma_{1}$) in this plane and denote the image by  \n\\[\nV_{0}=(p,0,-p).\n\\]\nPlace the vertex $V_{0}$ of the moving paraboloid (now called $\\Sigma_{2}$) at this point and orient $\\Sigma_{2}$ so that  \n\n\\qquad$\\bullet$ it is tangent to $\\Sigma_{1}$ at $Q_{0}$, and  \n\n\\qquad$\\bullet$ its symmetry axis $A_{2}(0)$ lies in the plane $y=0$,  \n\nso that $\\Sigma_{2}$ is entirely outside $\\Sigma_{1}$.  \nThe motion is then continued, still satisfying the no–slip/no–twist constraints, until $\\Sigma_{2}$ has returned to its initial attitude after one complete revolution about the $z$–axis.\n\n(a)  For every instant $t$ let $\\Pi(t)$ be the unique meridian plane containing the $z$–axis and the current contact point $Q(t)$.  \n\n(i) Prove that the symmetry axis $A_{2}(t)$ of $\\Sigma_{2}$ is contained in $\\Pi(t)$.  \n\n(ii) Let $\\boldsymbol\\omega(t)$ be the angular–velocity vector of $\\Sigma_{2}$ relative to $\\Sigma_{1}$.  Show that  \n\n\\[\n\\boldsymbol\\omega(t)\\in\\Pi(t)\\quad\\Longleftrightarrow\\quad\n\\bigl(\\forall P\\in\\Pi(t)\\bigr)\\,v_{P}(t)\\perp\\Pi(t),\n\\]\n\nand, making full use of the rolling constraints and the symmetries of the configuration, deduce that  \n\\[\n\\boxed{\\boldsymbol\\omega(t)\\in\\Pi(t)\\quad\\text{and}\\quad\n       \\boldsymbol\\omega(t)\\parallel\\mathbf e_{r}(t)},\n\\]\nthat is, $\\boldsymbol\\omega(t)$ is parallel to the radial principal direction.\n\n(b)  Restrict the study to the fixed meridian plane $\\Pi(0)=\\{y=0\\}$.  While the contact point $Q(t)$ remains in this plane, determine the planar locus $\\ell$ traced by the vertex $V(t)$ of $\\Sigma_{2}$ under the rolling constraints.  Show further that, when $\\Pi(t)$ is allowed to rotate with the motion, the full three–dimensional locus $\\mathcal L$ traced by $V$ during one complete revolution is the cubic surface  \n\n\\[\nF(x,y,z):=(x^{2}+y^{2}+z^{2})\\,z+2p\\,(x^{2}+y^{2})=0.\n\\]\n\nState explicitly which branch of $\\mathcal L$ is actually generated by the motion and justify every geometric step that leads from the planar curve $\\ell$ to the surface $F=0$.\n\n(c)  Determine the algebraic degree of $\\mathcal L$ and verify it directly from the equation $F=0$.",
      "solution": "Throughout boldface symbols denote vectors in $\\mathbb R^{3}$, ordinary letters their\ncoordinates in the fixed frame $(O;x,y,z)$.  \nAt the contact point $Q(t)$ put  \n\\[\n\\mathbf n(t)\\;\\text{(unit normal pointing outward from }\\Sigma_{1}),\\quad\n\\mathbf e_{r}(t)\\;\\text{(radial unit vector in }T_{Q(t)}\\Sigma_{1}),\\quad\n\\mathbf e_{\\theta}(t)=\\mathbf n(t)\\times\\mathbf e_{r}(t).\n\\]\nThe ordered triple $\\bigl(\\mathbf e_{r},\\mathbf e_{\\theta},\\mathbf n\\bigr)$ is direct and orthonormal.\n\n-----------------------------------------------------------------\n1.  Part (a)(i): the axis $A_{2}(t)$ lies in $\\Pi(t)$  \n-----------------------------------------------------------------\n\nExactly as in any surface–of–revolution argument, $\\mathbf e_{\\theta}(t)$ is the \\emph{unique} azimuthal principal direction at $Q(t)$.  \nBecause the no–twist constraint forces the principal frames of the two\ncongruent surfaces to coincide, the meridian direction of $\\Sigma_{2}(t)$ must\nbe the line orthogonal to $\\mathbf e_{\\theta}(t)$, namely $\\Pi(t)$.  Since the\nsymmetry axis of a surface of revolution belongs to each of its meridian\nplanes, we obtain\n\\[\n\\boxed{A_{2}(t)\\subset\\Pi(t)\\quad\\forall\\,t}.\n\\]\n\n-----------------------------------------------------------------\n2.  Part (a)(ii): direction of the angular–velocity vector  \n-----------------------------------------------------------------\n\n-----------------------------------------------------------------\n2.1  A correct proof of the key equivalence  \n-----------------------------------------------------------------  \n\nLet $\\Pi(t)$ be a plane through the origin\\footnote{Only the direction of the plane matters for the present argument; translating it by the contact point would not change anything.} with unit normal $\\mathbf m(t)=\\mathbf e_{\\theta}(t)$.  \nDecompose an arbitrary vector $\\boldsymbol\\omega$ into its normal and tangential components:\n\\[\n\\boldsymbol\\omega=\\omega_{\\perp}\\,\\mathbf m+\\boldsymbol\\omega_{\\parallel},\n\\qquad\\boldsymbol\\omega_{\\parallel}\\in\\Pi(t).\n\\]\nFor every $\\mathbf u\\in\\Pi(t)$ the linear velocity of the material point whose (instantaneous) position vector is $\\mathbf u$ reads\n\\[\n\\mathbf v(\\mathbf u)=\\boldsymbol\\omega\\times\\mathbf u.\n\\]\nTake any $\\mathbf w\\in\\Pi(t)$.  Using the scalar triple product,\n\\[\n\\mathbf v(\\mathbf u)\\cdot\\mathbf w\n       =(\\boldsymbol\\omega\\times\\mathbf u)\\cdot\\mathbf w\n       =\\boldsymbol\\omega\\cdot(\\mathbf u\\times\\mathbf w).\n\\]\nBecause $\\mathbf u,\\mathbf w\\in\\Pi(t)$, their cross–product $\\mathbf u\\times\\mathbf w$ is parallel to $\\mathbf m$.  Hence\n\\[\n\\mathbf v(\\mathbf u)\\cdot\\mathbf w\n       =\\omega_{\\perp}\\,\\mathbf m\\cdot(\\mathbf u\\times\\mathbf w).\n\\]\nTherefore  \n\n$\\displaystyle\\bigl(\\forall\\mathbf u,\\mathbf w\\in\\Pi(t)\\bigr)\\;\n         \\mathbf v(\\mathbf u)\\cdot\\mathbf w=0\n\\quad\\Longleftrightarrow\\quad\n         \\omega_{\\perp}=0\n\\quad\\Longleftrightarrow\\quad\n         \\boldsymbol\\omega\\in\\Pi(t).$\n\nIn words,\n\\[\n\\boxed{\\;\n\\boldsymbol\\omega\\in\\Pi(t)\n\\;\\Longleftrightarrow\\;\n\\bigl(\\forall P\\in\\Pi(t)\\bigr)\\,v_{P}\\perp\\Pi(t)\n\\;}\n\\tag{2.1}\n\\]\nwhich is the desired equivalence.\n\n-----------------------------------------------------------------\n2.2  Elimination of the normal component  \n-----------------------------------------------------------------\n\nThe no–twist constraint literally says that the relative spin about the common\nnormal is zero, so that  \n\\[\n\\omega_{n}:=\\boldsymbol\\omega\\cdot\\mathbf n(t)=0. \\tag{2.2}\n\\]\nTherefore $\\boldsymbol\\omega$ already belongs to $T_{Q(t)}\\Sigma_{1}$.  It\nremains to rule out a component along $\\mathbf e_{\\theta}(t)$.\n\n-----------------------------------------------------------------\n2.3  A symmetry argument that forces $\\omega_{\\theta}=0$  \n-----------------------------------------------------------------\n\nThe entire configuration at time $t$ is\n\\emph{mirror–symmetric} with respect to $\\Pi(t)$:\n\n(i)  The fixed paraboloid $\\Sigma_{1}$ is invariant under the reflection  \n\\[\n\\mathscr R_{\\Pi(t)}\\colon(x,y,z)\\longmapsto(x,-y,z).\n\\]\n\n(ii)  $\\Sigma_{2}(t)$ shares that symmetry, because its\nvertex $V(t)$ and its axis $A_{2}(t)$ both lie in $\\Pi(t)$ (proved in\npart (a)(i)).\n\n(iii)  The non–holonomic constraints (no–slip, no–twist) are geometric\nconditions preserved by every ambient isometry, in particular by\n$\\mathscr R_{\\Pi(t)}$.\n\nHence, if $\\bigl(\\Sigma_{2}(t),\\boldsymbol\\omega(t)\\bigr)$ is an admissible\nstate, so is  \n$\\bigl(\\mathscr R_{\\Pi(t)}\\Sigma_{2}(t),\\,\n       \\mathscr R_{\\Pi(t)\\,*}\\boldsymbol\\omega(t)\\bigr)$.\nBut by (ii) the first components coincide, therefore\n\\[\n\\mathscr R_{\\Pi(t)\\,*}\\boldsymbol\\omega(t)=\\boldsymbol\\omega(t).\n\\tag{2.3}\n\\]\n\nNow $\\mathscr R_{\\Pi(t)}$ fixes $\\mathbf e_{r}(t)$ and $\\mathbf n(t)$ but\n\\emph{reverses the sign} of $\\mathbf e_{\\theta}(t)$.  Writing\n\\[\n\\boldsymbol\\omega=\\omega_{r}\\,\\mathbf e_{r}\n                 +\\omega_{\\theta}\\,\\mathbf e_{\\theta},\n\\tag{2.4}\n\\]\nrelation \\eqref{2.3} gives\n\\[\n\\mathscr R_{\\Pi(t)\\,*}\\boldsymbol\\omega\n     =\\omega_{r}\\,\\mathbf e_{r}-\\omega_{\\theta}\\,\\mathbf e_{\\theta}\n     =\\boldsymbol\\omega\n     \\;\\Longrightarrow\\;\n     \\boxed{\\omega_{\\theta}=0}. \\tag{2.5}\n\\]\n\n-----------------------------------------------------------------\n2.4  Conclusion for the angular velocity  \n-----------------------------------------------------------------\n\nCombining \\eqref{2.2} and \\eqref{2.5},\n\\[\n\\boxed{\\boldsymbol\\omega(t)=\\omega_{r}(t)\\,\\mathbf e_{r}(t)},\\qquad\n\\boldsymbol\\omega(t)\\in\\Pi(t)\\;\\text{and}\\;\n\\boldsymbol\\omega(t)\\perp\\mathbf e_{\\theta}(t),\n\\]\nwhich simultaneously proves the boxed statement required in part (a)(ii).\n\n-----------------------------------------------------------------\n3.  Geometry inside the fixed meridian plane $\\Pi(0)=\\{y=0\\}$  \n-----------------------------------------------------------------\n\nSuppress the $y$–coordinate and rename $x=r,\\;z=z$.  \nIntersecting the paraboloids with the plane gives the congruent parabolas  \n\\[\nC_{1}\\colon\\;z=\\dfrac{r^{2}}{4p},\\qquad\nC_{2}\\colon\\;z=-\\dfrac{r^{2}}{4p}. \\tag{3.1}\n\\]\n\nParameterise the fixed parabola $C_{1}$ by the focal parameter $t>0$:\n\\[\nQ(t)=(r(t),z(t))=(2pt,\\;pt^{2}). \\tag{3.2}\n\\]\nThe tangent line $\\tau(t)$ at $Q(t)$ has slope $dz/dr=r/(2p)=t$; hence  \n\\[\n\\tau(t)\\colon\\;z-pt^{2}=t\\,(r-2pt). \\tag{3.3}\n\\]\n\nBecause the parabolas are congruent, $\\tau(t)$ is the axis of\nsymmetry for the ordered pair $(C_{1},C_{2})$.  Consequently the vertex\n$v(t)$ of $C_{2}$ is the reflection of the origin $O=(0,0)$ in $\\tau(t)$.\n\nFoot of the perpendicular.  \nLet $H(t)$ be the foot of the perpendicular from $O$ onto $\\tau(t)$.  Solving\n\\eqref{3.3} together with $z=-(1/t)\\,r$ yields  \n\\[\nH(t)=\\Bigl(\\dfrac{pt^{3}}{1+t^{2}},\\;-\\dfrac{pt^{2}}{1+t^{2}}\\Bigr). \\tag{3.4}\n\\]\n\nReflection formula.  \nTherefore  \n\\[\nv(t)=2H(t)-O\n     =\\Bigl(\\dfrac{2pt^{3}}{1+t^{2}},\\;-\\dfrac{2pt^{2}}{1+t^{2}}\\Bigr). \\tag{3.5}\n\\]\n\nAt $t=1$ one finds $v(1)=(p,-p)$, agreeing with the prescribed initial vertex\n$V_{0}$.\n\n-----------------------------------------------------------------\n4.  Implicit equation of the planar locus $\\ell$  \n-----------------------------------------------------------------\n\nEliminate $t$ from \\eqref{3.5}.  From the first component one gets\n$t=r_{v}/(-z_{v})$ (because $z_{v}<0$).  Substituting into the second\ncomponent gives  \n\\[\nz_{v}\\,(r_{v}^{2}+z_{v}^{2})+2p\\,r_{v}^{2}=0. \\tag{4.1}\n\\]\nThus  \n\\[\n\\boxed{\\ \\ell=\\bigl\\{(r,z)\\in\\mathbb R^{2}\\colon z(r^{2}+z^{2})+2p\\,r^{2}=0\\bigr\\}\\ }. \\tag{4.2}\n\\]\n\nPhysical branch.  \nDuring the motion $t>0\\;\\Rightarrow\\;r_{v}>0,\\;z_{v}<0$; hence only the branch\n$z<0$ of the cubic \\eqref{4.2} is realised.\n\n-----------------------------------------------------------------\n5.  Generation of the spatial locus $\\mathcal L$  \n-----------------------------------------------------------------\n\nBy part (a)(i) the contact point $Q(t)$ and the vertex $V(t)$ lie in the\n\\emph{same} meridian plane $\\Pi(t)$.  Consequently they share the same\nazimuthal angle $\\varphi(t)$.  Rotating the planar curve $\\ell$ about the\n$z$–axis by that angle sweeps the full spatial locus:\n\\[\n\\mathcal L=\\bigl\\{(r\\cos\\varphi,\\;r\\sin\\varphi,\\;z)\\colon\\;(r,z)\\in\\ell,\\;\n             \\varphi\\in[0,2\\pi)\\bigr\\}.\n\\]\nReplacing $r$ by $\\sqrt{x^{2}+y^{2}}$ in \\eqref{4.1} gives the Cartesian\nimplicit polynomial  \n\\[\nF(x,y,z):=(x^{2}+y^{2}+z^{2})\\,z+2p\\,(x^{2}+y^{2})=0,\n\\]\nwhich is exactly the surface announced in the problem.\n\nBranch realised.  \nBecause the motion keeps $z<0$, the physically attainable part is  \n\\[\n\\mathcal L_{\\text{phys}}=\\mathcal L\\cap\\{\\,z<0\\,\\}.\n\\]\n(The algebraic point $(0,0,0)$ satisfies $F=0$ but is never reached.)\n\n-----------------------------------------------------------------\n6.  Part (c): algebraic degree of $\\mathcal L$  \n-----------------------------------------------------------------\n\nThe polynomial $F$ contains the monomials $z^{3}$ and $z(x^{2}+y^{2})$, both\nof total degree $3$, and no term of higher degree.  Hence $\\mathcal L$ is a\ncubic surface of revolution.  Its intersection with the plane $y=0$ reproduces\n\\eqref{4.2}, confirming the degree once more.\n\n-----------------------------------------------------------------\n7.  Closure of the motion  \n-----------------------------------------------------------------\n\nWhen the azimuth $\\varphi(t)$ increases by $2\\pi$, the moving paraboloid\n$\\Sigma_{2}$ (hence its vertex) returns to its initial attitude; the branch\n$z<0$ of the cubic surface is therefore described exactly once during a full\nrevolution.\n\n-----------------------------------------------------------------\nAnswer.  \nThe vertex of the rolling paraboloid traces precisely the branch $z<0$ of the\ncubic surface  \n\\[\n\\boxed{(x^{2}+y^{2}+z^{2})\\,z+2p\\,(x^{2}+y^{2})=0}.\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.319823",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension – The original problem is planar; the enhanced variant takes place in 3-space, demanding the passage from a curve to a surface and an understanding of surfaces of revolution.\n\n2. Additional structures – Rolling contact between two 2-dimensional surfaces introduces differential-geometric ideas (common normal, no-slip kinematics) that are absent in the 2-D case.\n\n3. Reduction & symmetry arguments – A successful solution requires reducing the 3-D motion to a 2-D meridian problem via rotational symmetry, then lifting the answer back to 3-D.  Handling this reduction rigorously involves group actions and invariance considerations.\n\n4. More intricate algebra – Eliminating the parameter now produces a quartic surface equation instead of a cubic curve, and the algebraic manipulation is appreciably heavier.\n\n5. Conceptual depth – Understanding why the vertex is the reflection of O across the tangent in three dimensions (it is really the reflection in the tangent plane along the normal) calls for familiarity with reflections in planes and with tangent planes to surfaces.\n\nOverall, the enhanced kernel variant requires geometric insight in higher dimensions, facility with differential geometry of surfaces, exploitation of symmetry groups, and more elaborate algebraic elimination, making it substantially harder than both the original and the given kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}