{
  "index": "1974-A-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "ANA"
  ],
  "difficulty": "",
  "question": "A-5. Consider the two mutually tangent parabolas \\( y=x^{2} \\) and \\( y=-x^{2} \\). [These have foci at \\( (0,1 / 4) \\) and \\( (0,-1 / 4) \\), and directrices \\( y=-1 / 4 \\) and \\( y=1 / 4 \\), respectively.] The upper parabola rolls without slipping around the fixed lower parabola. Find the locus of the focus of the moving parabola.",
  "solution": "A-5.\nLet \\( F \\) be the fixed focus, \\( M \\) be the moving focus, and \\( T \\) be the (varying) point of mutual tangency. The reflecting property of parabolas tells us that the tangent line at \\( T \\) makes equal angles with \\( F T \\) and with a vertical line. This and congruence of the two parabolas imply that \\( M T \\) is vertical and that the segments \\( \\overline{F T} \\) and \\( \\overline{M T} \\) are equal. Now \\( M \\) must be on the horizontal fixed directrix \\( y=1 / 4 \\) by the focus-directrix definition of a parabola.",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "F",
    "M",
    "T"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "F": "fixedfocus",
        "M": "movingfocus",
        "T": "tangencypoint"
      },
      "question": "A-5. Consider the two mutually tangent parabolas \\( ordinate=abscissa^{2} \\) and \\( ordinate=-abscissa^{2} \\). [These have foci at \\( (0,1 / 4) \\) and \\( (0,-1 / 4) \\), and directrices \\( ordinate=-1 / 4 \\) and \\( ordinate=1 / 4 \\), respectively.] The upper parabola rolls without slipping around the fixed lower parabola. Find the locus of the focus of the moving parabola.",
      "solution": "A-5.\nLet \\( fixedfocus \\) be the fixed focus, \\( movingfocus \\) be the moving focus, and \\( tangencypoint \\) be the (varying) point of mutual tangency. The reflecting property of parabolas tells us that the tangent line at \\( tangencypoint \\) makes equal angles with \\( fixedfocus \\; tangencypoint \\) and with a vertical line. This and congruence of the two parabolas imply that \\( movingfocus \\; tangencypoint \\) is vertical and that the segments \\( \\overline{fixedfocus \\; tangencypoint} \\) and \\( \\overline{movingfocus \\; tangencypoint} \\) are equal. Now \\( movingfocus \\) must be on the horizontal fixed directrix \\( ordinate=1 / 4 \\) by the focus-directrix definition of a parabola."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "ironclasp",
        "y": "moonrider",
        "F": "bookshelf",
        "M": "driftwood",
        "T": "pineconed"
      },
      "question": "A-5. Consider the two mutually tangent parabolas \\( moonrider=ironclasp^{2} \\) and \\( moonrider=-ironclasp^{2} \\). [These have foci at \\( (0,1 / 4) \\) and \\( (0,-1 / 4) \\), and directrices \\( moonrider=-1 / 4 \\) and \\( moonrider=1 / 4 \\), respectively.] The upper parabola rolls without slipping around the fixed lower parabola. Find the locus of the focus of the moving parabola.",
      "solution": "A-5.\nLet \\( bookshelf \\) be the fixed focus, \\( driftwood \\) be the moving focus, and \\( pineconed \\) be the (varying) point of mutual tangency. The reflecting property of parabolas tells us that the tangent line at \\( pineconed \\) makes equal angles with \\( bookshelf\\;pineconed \\) and with a vertical line. This and congruence of the two parabolas imply that \\( driftwood\\;pineconed \\) is vertical and that the segments \\( \\overline{bookshelf\\;pineconed} \\) and \\( \\overline{driftwood\\;pineconed} \\) are equal. Now \\( driftwood \\) must be on the horizontal fixed directrix \\( moonrider=1 / 4 \\) by the focus-directrix definition of a parabola."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "F": "diffusion",
        "M": "stationary",
        "T": "separation"
      },
      "question": "A-5. Consider the two mutually tangent parabolas \\( horizontalaxis=verticalaxis^{2} \\) and \\( horizontalaxis=-verticalaxis^{2} \\). [These have foci at \\( (0,1 / 4) \\) and \\( (0,-1 / 4) \\), and directrices \\( horizontalaxis=-1 / 4 \\) and \\( horizontalaxis=1 / 4 \\), respectively.] The upper parabola rolls without slipping around the fixed lower parabola. Find the locus of the focus of the moving parabola.",
      "solution": "A-5.\nLet \\( diffusion \\) be the fixed focus, \\( stationary \\) be the moving focus, and \\( separation \\) be the (varying) point of mutual tangency. The reflecting property of parabolas tells us that the tangent line at \\( separation \\) makes equal angles with \\( diffusion separation \\) and with a vertical line. This and congruence of the two parabolas imply that \\( stationary separation \\) is vertical and that the segments \\( \\overline{diffusion separation} \\) and \\( \\overline{stationary separation} \\) are equal. Now \\( stationary \\) must be on the horizontal fixed directrix \\( horizontalaxis=1 / 4 \\) by the focus-directrix definition of a parabola."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "F": "vbmnlkey",
        "M": "sdjqopru",
        "T": "aghyzifw"
      },
      "question": "A-5. Consider the two mutually tangent parabolas \\( hjgrksla=qzxwvtnp^{2} \\) and \\( hjgrksla=-qzxwvtnp^{2} \\). [These have foci at \\( (0,1 / 4) \\) and \\( (0,-1 / 4) \\), and directrices \\( hjgrksla=-1 / 4 \\) and \\( hjgrksla=1 / 4 \\), respectively.] The upper parabola rolls without slipping around the fixed lower parabola. Find the locus of the focus of the moving parabola.",
      "solution": "A-5.\nLet \\( vbmnlkey \\) be the fixed focus, \\( sdjqopru \\) be the moving focus, and \\( aghyzifw \\) be the (varying) point of mutual tangency. The reflecting property of parabolas tells us that the tangent line at \\( aghyzifw \\) makes equal angles with \\( vbmnlkey aghyzifw \\) and with a vertical line. This and congruence of the two parabolas imply that \\( sdjqopru aghyzifw \\) is vertical and that the segments \\( \\overline{vbmnlkey aghyzifw} \\) and \\( \\overline{sdjqopru aghyzifw} \\) are equal. Now \\( sdjqopru \\) must be on the horizontal fixed directrix \\( hjgrksla=1 / 4 \\) by the focus-directrix definition of a parabola."
    },
    "kernel_variant": {
      "question": "Let $\\mathbf R^{3}$ carry the usual Cartesian frame $(x,y,z)$.  \nFix the upward–opening circular paraboloid  \n\\[\n\\Sigma_{0}\\;:\\; z=\\dfrac{x^{2}+y^{2}}{4p},\\qquad p>0,\n\\]\nwhose focus and directrix plane are, respectively,  \n\\[\nF_{0}=(0,0,p),\\qquad \\Pi_{0}\\;:\\; z=-p.\n\\]\n\nChoose a number $c_{0}>0$ and set  \n\\[\nR:=\\sqrt{2pc_{0}},\\qquad \nT_{0}:=(R,0,c_{0}/2).\n\\]\n(The point $T_{0}$ lies on $\\Sigma_{0}$ and will be the initial point of contact.)\n\nIntroduce the downward–opening paraboloid  \n\\[\n\\Sigma_{0}^{\\downarrow}\\;:\\; z=c_{0}-\\dfrac{(x-2R)^{2}+y^{2}}{4p}.\n\\]\nIt is congruent to $\\Sigma_{0}$, its axis is the $z$–axis, and it is tangent to $\\Sigma_{0}$ precisely at $T_{0}$.\n\nFor $t\\ge 0$ let a moving paraboloid $\\Sigma_{t}$ satisfy\n\n(C1)\\; $\\Sigma_{t}$ is always congruent to $\\Sigma_{0}$ and its axis is parallel to the $z$–axis (it may translate parallel to the $xy$–plane and rotate about the $z$–direction, but it never tilts);\n\n(C2)\\; $\\Sigma_{t}$ is tangent to $\\Sigma_{0}$ at exactly one point $T(t)$;\n\n(C3)\\; the relative motion of $\\Sigma_{t}$ with respect to the fixed $\\Sigma_{0}$ is, at every instant $t$, a {\\em pure rotation about the vertical line through $T(t)$}.  \nEquivalently, the instantaneous angular–velocity vector satisfies  \n$\\boldsymbol\\Omega(t)=\\omega(t)\\mathbf k$ (with $\\mathbf k$ the unit vector in the positive $z$–direction), and the material point of $\\Sigma_{t}$ that occupies $T(t)$ is momentarily at rest (so there is no slip at $T(t)$).\n\nInitially $\\Sigma_{t}\\bigl\\vert_{t=0}=\\Sigma_{0}^{\\downarrow}$.\n\nFor each $t\\ge 0$ denote  \n\n(i)\\; $M(t)=\\;$focus of $\\Sigma_{t}$,  \n\n(ii)\\; $\\Pi_{t}=\\;$directrix plane of $\\Sigma_{t}$.\n\n(a)\\; Prove that the vertex level $c_{t}$ of $\\Sigma_{t}$ is independent of $t$ and equals $c_{0}$, and determine explicitly the spatial locus  \n\\[\n\\Lambda=\\{\\,M(t):t\\ge 0\\,\\}.\n\\]\n\n(b)\\; Show that the distance $\\lvert F_{0}M(t)\\rvert$ is constant and equals $c_{0}+2p$.\n\n(c)\\; Compute the exact length of one complete circuit of $\\Lambda$ and obtain an explicit equation for the surface swept out by the family of directrix planes $\\{\\Pi_{t}\\}_{t\\ge 0}$.",
      "solution": "Throughout we use cylindrical coordinates  \n\\[\nx=r\\cos\\theta,\\quad y=r\\sin\\theta,\\quad r^{2}=x^{2}+y^{2}.\n\\tag{1}\n\\]\n\n1.\\;A general congruent copy whose axis is parallel to $\\mathbf k$  \nBecause $\\Sigma_{0}$ is a surface of revolution, every congruent copy whose axis is {\\em parallel} to $\\,\\mathbf k$ (not necessarily passing through the origin) possesses an equation of the form  \n\\[\n\\Sigma(a,b,c):\\; z=c-\\dfrac{(x-a)^{2}+(y-b)^{2}}{4p},\n\\qquad (a,b,c)\\in\\mathbf R^{3}.\n\\tag{2}\n\\]\nIts focus and directrix are  \n\\[\nM=(a,b,c-p),\\qquad \n\\Pi:\\;z=c+p.\n\\tag{3}\n\\]\nHence three scalar functions $(a_{t},b_{t},c_{t})$ completely determine $\\Sigma_{t}$; an additional rotation about the $z$–direction is immaterial to (2)–(3).\n\n2.\\;Consequences of the single tangency (C2)  \nLet  \n\\[\nT(t)=(x,y,z),\\qquad r=\\sqrt{x^{2}+y^{2}}>0.\n\\tag{4}\n\\]\nBecause $T(t)$ lies on both $\\Sigma_{0}$ and $\\Sigma_{t}$,\n\\[\nz=\\dfrac{r^{2}}{4p}=c_{t}-\\dfrac{(x-a_{t})^{2}+(y-b_{t})^{2}}{4p}.\n\\tag{5}\n\\]\nEquality of tangent planes at $T(t)$ means their (unnormalised) normals  \n\\[\nn_{0}=\\Bigl(-\\dfrac{x}{2p},-\\dfrac{y}{2p},1\\Bigr),\\qquad\nn_{t}=\\Bigl(\\dfrac{x-a_{t}}{2p},\\dfrac{y-b_{t}}{2p},1\\Bigr)\n\\tag{6}\n\\]\nare parallel.  The third components already agree, hence the proportionality factor must be $1$, giving  \n\\[\n(x-a_{t},\\,y-b_{t})=(-x,-y)\\;\\Longrightarrow\\;\na_{t}=2x,\\;b_{t}=2y.\n\\tag{7}\n\\]\nConsequently the horizontal projection of $T(t)$ is the midpoint of the segment $OM(t)$.  \nSubstituting (7) into (5) yields  \n\\[\n\\dfrac{r^{2}}{4p}=c_{t}-\\dfrac{r^{2}}{4p}\\quad\\Longrightarrow\\quad\nr^{2}=2pc_{t}.\n\\tag{8}\n\\]\n\n3.\\;Rolling without slipping $\\Longrightarrow\\;\\dot c_{t}\\equiv 0$  \nLet\n\\[\nV_{t}=(a_{t},b_{t},c_{t})\n\\tag{9}\n\\]\nbe the vertex of $\\Sigma_{t}$.  \nFrom (7)–(8) together with $z=r^{2}/(4p)$ one has  \n\\[\nT=(x,y,z)=\\Bigl(x,y,\\dfrac{c_{t}}{2}\\Bigr),\\qquad\nV_{t}-T=(x,y,c_{t}/2).\n\\tag{10}\n\\]\n\nBy (C3) the instantaneous angular velocity of $\\Sigma_{t}$ relative to $\\Sigma_{0}$ is  \n\\[\n\\boldsymbol\\Omega=\\omega\\,\\mathbf k,\\qquad \\omega\\in\\mathbf R.\n\\tag{11}\n\\]\nFor any point $Q\\in\\Sigma_{t}$ the spatial velocity is  \n\\[\n\\mathbf v_{Q}=\\boldsymbol\\Omega\\times(Q-T).\n\\tag{12}\n\\]\nIn particular,\n\\[\n\\mathbf v_{V}\n      =\\boldsymbol\\Omega\\times\\bigl(V_{t}-T\\bigr)\n      =\\omega\\,\\mathbf k\\times(x,y,c_{t}/2)\n      =\\bigl(-\\omega y,\\;\\omega x,\\;0\\bigr).\n\\tag{13}\n\\]\nThe $z$–component vanishes, so the vertex possesses no vertical velocity.  Therefore  \n\\[\n\\dot c_{t}=0\\quad\\forall\\,t\\ge 0,\n\\qquad\\Longrightarrow\\qquad c_{t}\\equiv c_{0}.\n\\tag{14}\n\\]\n\n4.\\;Geometry of the tangency circle and the focus locus $\\Lambda$  \nBecause $c_{t}=c_{0}$, identity (8) gives $r^{2}=2pc_{0}$; hence $T(t)$ travels around the fixed horizontal circle  \n\\[\nT(\\theta)=\\bigl(R\\cos\\theta,\\;R\\sin\\theta,\\;c_{0}/2\\bigr),\n\\qquad 0\\le\\theta<2\\pi,\n\\tag{15}\n\\]\nwhere $R=\\sqrt{2pc_{0}}$.  \nUsing (3), (7) and $c_{t}=c_{0}$ we obtain  \n\\[\nM(\\theta)=\\bigl(2R\\cos\\theta,\\;2R\\sin\\theta,\\;c_{0}-p\\bigr).\n\\tag{16}\n\\]\nThus  \n\\[\n\\Lambda=\\{(x,y,z)\\in\\mathbf R^{3}:\\;x^{2}+y^{2}=4R^{2}=8pc_{0},\\;z=c_{0}-p\\},\n\\tag{17}\n\\]\na horizontal circle of radius $\\,2\\sqrt{2pc_{0}}\\,$ situated in the plane $z=c_{0}-p$.  \nThis completes part (a).\n\n5.\\;Constancy of $\\lvert F_{0}M(t)\\rvert$  \nWith $F_{0}=(0,0,p)$ and (16),\n\\[\n\\lvert F_{0}M\\rvert^{2}\n   =(2R)^{2}+\\,\\bigl[(c_{0}-p)-p\\bigr]^{2}\n   =4(2pc_{0})+\\bigl(c_{0}-2p\\bigr)^{2}\n   =(c_{0}+2p)^{2},\n\\tag{18}\n\\]\nso $\\lvert F_{0}M(t)\\rvert=c_{0}+2p$ is indeed constant, as required in part (b).\n\n6.\\;Length of one circuit of $\\Lambda$  \nEquation (17) shows that $\\Lambda$ is a circle of radius  \n\\[\nR_{\\Lambda}=2\\sqrt{2pc_{0}},\n\\tag{19}\n\\]\nwhence  \n\\[\n\\operatorname{len}(\\Lambda)=2\\pi R_{\\Lambda}=4\\pi\\sqrt{2pc_{0}}.\n\\tag{20}\n\\]\n\n7.\\;Surface swept out by the directrix planes $\\Pi_{t}$  \nBecause $c_{t}\\equiv c_{0}$, formula (3) gives  \n\\[\n\\Pi_{t}:\\;z=c_{0}+p,\n\\tag{21}\n\\]\nindependent of $t$.  Thus the family $\\{\\Pi_{t}\\}_{t\\ge 0}$ coincides with the single horizontal plane  \n\\[\nz=c_{0}+p.\n\\tag{22}\n\\]\n\nStatements (17), (20) and (22) finish part (c). \\qed",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.613243",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the problem moves from planar curves to 3-dimensional surfaces (paraboloids).  \n• Additional interacting conditions: tangency, rolling without slipping, and preservation of the vertical axis must all be handled simultaneously.  \n• Deeper theory: the solution requires differential-geometry tools (surface normals, tangent–plane calculations, parallelism of gradients) together with classical focal/directrix properties.  \n• Multiple results: the solver must find a locus {\\it and} prove an orthogonality/length property, then compute both a curve length and a swept surface.  \n• Non-trivial algebra: eliminating parameters to obtain (3) and establishing the constant length in Step 6 are considerably more involved than the 2-dimensional equal-segment argument of the original problem.\n\nAll these layers make the enhanced variant significantly more technical and conceptually demanding than both the original problem and the existing kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\mathbf R^{3}$ carry the usual Cartesian coordinates $(x,y,z)$.  Fix the upward–opening circular paraboloid  \n\\[\n\\Sigma_{0}\\;:\\; z=\\dfrac{x^{2}+y^{2}}{4p},\\qquad p>0,\n\\]\nwhose focus and directrix are, respectively,  \n\\[\nF_{0}=(0,0,p),\\qquad\\Pi_{0}\\;:\\; z=-p.\n\\]\n\nChoose a number $c_{0}>0$ and put  \n\\[\nR:=\\sqrt{2pc_{0}}. \\tag{1}\n\\]\nDenote  \n\\[\nT_{0}:=(R,0,c_{0}/2). \\tag{2}\n\\]\n\nConsider the downward–opening paraboloid  \n\\[\n\\Sigma_{0}^{\\downarrow}\\;:\\; z=c_{0}-\\dfrac{(x-2R)^{2}+y^{2}}{4p}. \\tag{3}\n\\]\nIt is congruent to $\\Sigma_{0}$, has vertical symmetry axis, and is tangent to $\\Sigma_{0}$ precisely at $T_{0}$.\n\nFor $t\\ge 0$ let a moving paraboloid $\\Sigma_{t}$ satisfy the following:\n\n(C1) $\\Sigma_{t}$ is always congruent to $\\Sigma_{0}$ and its symmetry axis is the $z$–axis (it may translate parallel to the $xy$–plane, translate vertically, and rotate about the $z$–axis, but it never tilts);\n\n(C2) $\\Sigma_{t}$ is tangent to $\\Sigma_{0}$ at exactly one point $T(t)$;\n\n(C3) the relative motion of $\\Sigma_{t}$ with respect to $\\Sigma_{0}$ is, at every instant $t$, a {\\em pure rotation about the common normal line through $T(t)$}.  In particular, the material point of $\\Sigma_{t}$ that occupies $T(t)$ is momentarily at rest in space (there is no slip at $T(t)$).\n\nInitially $\\Sigma_{0}^{\\downarrow}$ is taken for $\\Sigma_{t}$ at $t=0$.\n\nFor every $t\\ge 0$ define  \n\\[\n\\text{(i) }M(t)=\\text{focus of }\\Sigma_{t},\\qquad\n\\text{(ii) }\\Pi_{t}=\\text{directrix plane of }\\Sigma_{t}.\n\\]\n\n(a)  Prove that the vertex level $c_{t}$ of $\\Sigma_{t}$ is independent of $t$ and equals $c_{0}$, and determine explicitly the spatial locus  \n\\[\n\\Lambda=\\{\\,M(t):t\\ge 0\\,\\}.\n\\]\n\n(b)  Show that the distance $\\lvert F_{0}M(t)\\rvert$ is constant and equals $c_{0}+2p$.\n\n(c)  Compute the exact length of one complete circuit of $\\Lambda$ and obtain an explicit equation for the surface swept out by the family of directrix planes $\\{\\Pi_{t}\\}_{t\\ge 0}$.",
      "solution": "Throughout we employ cylindrical coordinates  \n\\[\nx=r\\cos\\theta,\\qquad y=r\\sin\\theta,\\qquad r^{2}=x^{2}+y^{2}. \\tag{4}\n\\]\n\n1.\\;A general congruent copy of $\\Sigma_{0}$ whose axis is the $z$–axis  \nBecause $\\Sigma_{0}$ is a surface of revolution, every such copy has an equation of the form  \n\\[\n\\Sigma(a,b,c):\\; z=c-\\dfrac{(x-a)^{2}+(y-b)^{2}}{4p},\\qquad c\\in\\mathbf R, \\tag{5}\n\\]\nwith focus and directrix  \n\\[\nM=(a,b,c-p),\\qquad \\Pi:\\;z=c+p. \\tag{6}\n\\]\nThus three scalar functions $(a_{t},b_{t},c_{t})$ completely specify $\\Sigma_{t}$; an extra rotation about the $z$–axis is immaterial to \\eqref{5}–\\eqref{6}.\n\n2.\\;Consequences of the single tangency (C2)  \nLet  \n\\[\nT(t)=(x,y,z),\\qquad r=\\sqrt{x^{2}+y^{2}}>0. \\tag{7}\n\\]\nSince $T(t)$ lies on both $\\Sigma_{0}$ and $\\Sigma_{t}$,  \n\\[\nz=\\dfrac{r^{2}}{4p}=c_{t}-\\dfrac{(x-a_{t})^{2}+(y-b_{t})^{2}}{4p}. \\tag{8}\n\\]\nBecause their tangent planes coincide at $T(t)$, their (un–normalised) normals  \n\\[\nn_{0}=\\Bigl(-\\dfrac{x}{2p},-\\dfrac{y}{2p},1\\Bigr),\\qquad\nn_{t}=\\Bigl(\\dfrac{x-a_{t}}{2p},\\dfrac{y-b_{t}}{2p},1\\Bigr) \\tag{9}\n\\]\nmust be parallel.  Equality of the third components forces the proportionality factor to be $1$, whence  \n\\[\n(x-a_{t},\\,y-b_{t})=(-x,-y)\\;\\Longrightarrow\\; a_{t}=2x,\\;b_{t}=2y. \\tag{10}\n\\]\nThus the horizontal projection of $T(t)$ is the midpoint of the segment $OM(t)$.  Substituting \\eqref{10} into \\eqref{8} yields  \n\\[\n\\dfrac{r^{2}}{4p}=c_{t}-\\dfrac{r^{2}}{4p}\\;\\Longrightarrow\\; r^{2}=2pc_{t}. \\tag{11}\n\\]\n\n3.\\;Rolling without slipping $\\Longrightarrow \\dot c_{t}\\equiv 0$  \nIntroduce the vertex  \n\\[\nV_{t}=(a_{t},\\,b_{t},\\,c_{t}). \\tag{12}\n\\]\nFrom \\eqref{10}–\\eqref{11} and $z=r^{2}/(4p)$ one has  \n\\[\nT=(x,y,z)=\\Bigl(x,y,\\dfrac{c_{t}}{2}\\Bigr),\\qquad\nV_{t}-T=(x,y,c_{t}/2). \\tag{13}\n\\]\n\nCondition (C3) states that at the instant $t$ the rigid body $\\Sigma_{t}$ performs a {\\em pure rotation} about the common normal line through $T$.  Denote the instantaneous angular velocity by  \n\\[\n\\boldsymbol\\Omega=\\omega\\,\\mathbf N,\\qquad \\omega\\in\\mathbf R,\\quad\n\\mathbf N=\\text{unit normal at }T\\text{ to }\\Sigma_{0}\\ (\\text{hence to }\\Sigma_{t}). \\tag{14}\n\\]\nFor any point $Q\\in\\Sigma_{t}$ the spatial velocity is  \n\\[\n\\mathbf v_{Q}=\\boldsymbol\\Omega\\times(Q-T). \\tag{15}\n\\]\nIn particular, the vertex velocity is  \n\\[\n\\mathbf v_{V}=\\boldsymbol\\Omega\\times(V_{t}-T). \\tag{16}\n\\]\n\nChoose the un–normalised normal $n_{0}$ of \\eqref{9},  \n\\[\nn_{0}=\\Bigl(-\\dfrac{x}{2p},-\\dfrac{y}{2p},1\\Bigr). \\tag{17}\n\\]\nBecause any scalar multiple of $\\boldsymbol\\Omega$ gives the same velocity field, we may take $\\boldsymbol\\Omega=n_{0}$.  Using \\eqref{13} we compute  \n\\[\n\\boldsymbol\\Omega\\times(V_{t}-T)=\n\\begin{vmatrix}\n\\mathbf i & \\mathbf j & \\mathbf k\\\\[2pt]\n-\\dfrac{x}{2p} & -\\dfrac{y}{2p} & 1\\\\[6pt]\nx & y & \\dfrac{c_{t}}{2}\n\\end{vmatrix}.\n\\]\nIts $k$–component equals  \n\\[\n-\\dfrac{x}{2p}\\,y-\\Bigl(-\\dfrac{y}{2p}\\Bigr)x=0. \\tag{18}\n\\]\nHence $(\\mathbf v_{V})_{z}=0$: the vertex has no vertical velocity, so  \n\\[\n\\dot c_{t}=0\\quad\\forall\\,t,\\qquad\\Longrightarrow\\qquad c_{t}\\equiv c_{0}. \\tag{19}\n\\]\n\n4.\\;Geometry of the tangency circle and of the focus locus $\\Lambda$  \nBecause $c_{t}$ is constant, \\eqref{11} gives $r\\equiv R=\\sqrt{2pc_{0}}$, and $T(t)$ runs along the fixed horizontal circle  \n\\[\nT(\\theta)=\\bigl(R\\cos\\theta,\\,R\\sin\\theta,\\,c_{0}/2\\bigr),\\qquad 0\\le\\theta<2\\pi. \\tag{20}\n\\]\nFrom \\eqref{6}, \\eqref{10}, and \\eqref{19},  \n\\[\nM(\\theta)=\\bigl(2R\\cos\\theta,\\,2R\\sin\\theta,\\,c_{0}-p\\bigr). \\tag{21}\n\\]\nTherefore  \n\\[\n\\Lambda=\\bigl\\{(x,y,z)\\in\\mathbf R^{3}:\\;x^{2}+y^{2}=4R^{2}=8pc_{0},\\;z=c_{0}-p\\bigr\\}. \\tag{22}\n\\]\nThus $\\Lambda$ is the horizontal circle of radius $2\\sqrt{2pc_{0}}$ lying in the plane $z=c_{0}-p$, completing part (a).\n\n5.\\;Constancy of $\\lvert F_{0}M(t)\\rvert$ (part (b))  \nWith $F_{0}=(0,0,p)$ and \\eqref{21},  \n\\[\n\\lvert F_{0}M\\rvert^{2}=(2R)^{2}+(c_{0}-p-p)^{2}\n=4R^{2}+(c_{0}-2p)^{2}\n=8pc_{0}+c_{0}^{2}-4pc_{0}+4p^{2}=(c_{0}+2p)^{2}. \\tag{23}\n\\]\nHence $\\lvert F_{0}M(t)\\rvert=c_{0}+2p$ is constant, proving part (b).\n\n6.\\;Length of one circuit of $\\Lambda$ (part (c))  \nEquation \\eqref{22} shows that $\\Lambda$ is a circle of radius  \n\\[\nR_{\\Lambda}=2\\sqrt{2pc_{0}}. \\tag{24}\n\\]\nConsequently  \n\\[\n\\operatorname{length}(\\Lambda)=2\\pi R_{\\Lambda}=4\\pi\\sqrt{2pc_{0}}. \\tag{25}\n\\]\n\n7.\\;Surface swept by the directrix planes $\\Pi_{t}$ (part (c))  \nBecause $c_{t}\\equiv c_{0}$, formula \\eqref{6} yields  \n\\[\n\\Pi_{t}:\\; z=c_{0}+p,\\qquad\\text{independent of }t. \\tag{26}\n\\]\nHence the entire family $\\{\\Pi_{t}\\}_{t\\ge 0}$ coincides with the single horizontal plane  \n\\[\nz=c_{0}+p. \\tag{27}\n\\]\n\nEquations \\eqref{22}, \\eqref{25}, and \\eqref{27} complete part (c). \\qed",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.490763",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the problem moves from planar curves to 3-dimensional surfaces (paraboloids).  \n• Additional interacting conditions: tangency, rolling without slipping, and preservation of the vertical axis must all be handled simultaneously.  \n• Deeper theory: the solution requires differential-geometry tools (surface normals, tangent–plane calculations, parallelism of gradients) together with classical focal/directrix properties.  \n• Multiple results: the solver must find a locus {\\it and} prove an orthogonality/length property, then compute both a curve length and a swept surface.  \n• Non-trivial algebra: eliminating parameters to obtain (3) and establishing the constant length in Step 6 are considerably more involved than the 2-dimensional equal-segment argument of the original problem.\n\nAll these layers make the enhanced variant significantly more technical and conceptually demanding than both the original problem and the existing kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}